1. To finde whether a Plain be Level or Horizontal.

The performance hereof is already shewed by Mr. Stirrup, in his Compleat Dyallist, Page 58. where he hath a Scheme to this purpose, I shall onely mention it: Get a smooth Board, let it have one right edge, and near to that edge, let a hole be cut in it for a Plummet to play in; draw a Line cross the Board Perpendicular, to the streight edge thereof passing through the former hole; if then setting the Board on its smooth edge, and holding it Perpen­dicularly, [Page 3] so that the Plummet may play in the hole, if which way soever the Board be turned, the threed will fall (being held up­right with the Plummet at the end of it, playing in the former hole) directly on the Perpendicular Line drawn cross the Board, the said Line no wayes Reclining from it, the Plain is Horizontal, otherwise not.

3. To take the Altitude of the Sun without Instrument.

Upon any Smooth Board, draw two Lines at right Angles, as AB and AC, and upon A as a Center with 60 ^d of a Line of Chords describe the Arch BC, and into the Center thereof, at A drive in a Pin or Steel Needle, upon which hang a threed and Plummet, then when you would observe an Altitude, hold the Board so to the Sun, that the shaddow of the Pin or Needle may fall on the Line AC, and where the threed intersects the Arch BC, set a mark, suppose at E then measure the Ark BE on your Line of Chords, and it shews the Altitude sought, for want of a Line of Chords, you may first divide the Arch BC into 9 parts, and the first of those Divisions into 10 smaller parts.

3. To draw a Horizontal and Vertical Line upon a Plain.

The readyest and most certainest way to do this, especially if the Plain lean downwards from the Zenith, will be by help of a threed and Plummet held steadily, to make two pricks at a compe­tent [Page 4] distance in the shadow of the threed on the Plain, projected by the eye, and a Line drawn through those, or parallel to those two points, shall be the Plains perpendicular or Vertical Line, and a Line drawn perpendicular to the said Line, shall be the Plains Horizontal Line.

To finde the Reclination of a Plain.

The Reclination of a Plain, is the Angle comprehended be­tween the Plains perpendicular, and the Axis of the Horizon, and the same Definition may serve for the Inclination, onely the upper face of a Plain, leaning from the Zenith, is said to recline, and the under face to incline. In this sense Mr. Oughtred in his Circles of Proportion, Mr. Wells in his Dyalling, Mr. Newton in his Institu­tion, and Mr. Foster in his late writings, understand it, and so it is to be taken throughout this Treatise; but here it will not be a­miss to intimate that Mr. Gunter, Mr. Wingate, and Mr. Foster in a Treatise of a Quadrant, published in Anno 1638. account the Inclination from the Horizon; the complement of the Angle here defined, and accordingly have suited their Proportions thereto, in which sense, the Reclining side is said to be the upper face of an Incliner; but for the future it will be inconvenient to take it any more in that Acception.

Now to finde the Reclination of a Plain, apply the streight edge of the Board we used for trying of Horizontal Plains, to the Plains perpendicular, holding the threed and plummet so in the grooved hole, that it may intersect the line, parallel to the streight edge of the Board, and make two pricks upon the Board where the threed pas­seth, and it will make an Angle with the said Line, equal to the Plains Reclination, to be measured with Chords, according as was directed for taking of Altitudes; and if need require, a Line may be drawn parallel to that drawn on the Board, which repre­sented the threed, which will make the same Angle with the Plains perpendicular, the former Line did.

The Inclination will be easily got, by applying the streight edge of the Board, to the Plains perpendicular, and holding the threed and plummet so at liberty, as that it may cross the Line parallel to the Boards streight edge, and it will make an Angle therewith, equal to the Plains Inclination.

Of Declining Plains.

A Plain hath its Denominations from the Scituation of its Poles: by the Poles of a Plain, is meant a Line imagined to pass through the Plain at right Angles thereto, the Extremities of which Line on both sides the Plain, are called its Poles. If a Plain look full South, without swerving, either to the East or West; it is said to be a direct South Plain, if it swerve towards East or West, it is said to decline thereto; if it stand upright without leaning, it is said to be erect Now a South Plain that declines Eastward, the oppo­site face thereto, is said to be a North Plain declining as much West­ward, by which means the Declination of a Plain will never exceed 90 ^d, and to which sense the following directions toroughout this Treatise are suited. Now we may define the Declination of a Plain to be the Arch of the Horizon, contained between the true Points of East or West, and the Plain, equal whereto is the Arch of the Ho­rizon between the true North or South, and that Vertical Circle or Azimuth that passeth through the Plains Poles.

To finde the Declination of a Plain.

Upon a Board, having one streight smooth edge, draw a Line pa­rallel thereto, and thereon describe a Semicircle, with the Radius of the Chords, which distinguish into two Quadrants by a perpendicu­lar from the Center; then holding the Board parallel to the Hori­zon, and applying the streight edge to the Wall, hold up a threed and Plummet, so that the shadow thereof may pass through the Center, and in the shadow, make a mark near the Limb, from which a Line drawn into the Center shal represent the shadow of the thread.

In this Scheam, let ☉ C represent the shadow of the threed passing tho­row the Center; at the same instant, take the Suns Alti­tude, and by the fol­lowing directions, finde the Suns true [Page 6] Azimuth. Admit the Sun be 75 ^d to the Eastwards of the South, prick it from ☉ to S, then the distance between P and S shews the quantity of the Plains Declination, and the Coast is also shewn, for if P fall to the Eastwards of S, the Declination is Eastwards, if to the Westwards of it, it is Westward; in this Example it is 30 ^d Eastwards.

We may suppose that the Reader knoweth which way from the Line of shadow to set of the Suns true Coast, that he may easily do by observing the Coast of his rising or setting, or whether his Al­titude increase or decrease, and needs no directions.

A General Proportion for finding the Azimuth.

Upon C as a Center, describe the Semicircle EBM, and draw ECM the Diameter, and CB from the Center perpendicular thereto, from B set off the Declination to D, the Latitude to L, the Altitude to A; and draw the Line CA continued. The nearest distance from D to BC when the Declination is North, place from C towards M when South towards E, and thereto set K, assuming the Diameter EM to represent the Secant of the Latitude, the Radius [Page 7] of the said Secant shall be the Cosine of the Latitude doubled; wherefore the nearest distance from L to CM doubled, shall be the Radius to the said Secant, which prick on the Line of Altitude from C to Z, then the nearest distance from Z to CM shall be the Co­sine of the Altitude to that Radius, which extent prick from E to F, and draw the Lines EF and FM; then for the third term of the proportion, take the distance between A and L, and prick it from M to G, from which point take the nearest distance to BC, and place it from C to X, so is the distance KX, the difference of the Ver­sed Sines sought, which prick from F to Q, and draw QR paral­lel to FM, so is RM the Versed Sine of the Suns Azimuth from the North sought, and CR the Sine of it to the Southwards of the East or West, and BP the measure thereof in the Limbe, found by drawing a LIne from R parallel to GB; in this Example the Suns true Goast is 15 ^d to the Southwards of the East or West. Note al­so, that the Point R may be found without drawing the Lines EF, QR by entring the extent KX, so that one foot resting on the Diameter, the other turned about, may but just touch FM.

To make a Horizontal Dyal.

A Plain is then said to be Horizontal, when it is parallel to the Horizontal Circle of the Sphere proper to the place of Habitation; these are such kinde of Dyals as stand commonly upon posts in a Garden.

By the former Directions, a true Meridian Line may be found, let it here be represented by the Line FM.

A Horizontal Dyal for the Latitude of London, 51 degrees, 32 minutes.

Cross the same with a perpendicular, to wit, the Line of Six, and draw an Oblong or square figure as here is done, wherein to write the hours, from M to C set off such a Radius as you intend to draw a Circle withall, which for convenience may be as big as the Plain will admit, and therewith upon C as a Center describe a Circle, and set off the Radius of the said Circle from F to d, from M to k, and draw the right Line Fd, then prick the Poles height or Latitude of the place from f to L, and from L to S, and through the point S draw the Line MS, and it shall represent the Cock or Stiles height [Page 8] from the point L, take the nearest distance to FM, and prick that extent from F to N on the Line Fd, then a Ruler laid from N to k, where it intersects the Meridian is the Regulating point, or point ☉, then from M divide the Circle into twelve equal parts for the whole hours, setting the Letters of the Alphabet thereto, and lay a ruler from the Regulating point to each of those Divisions, and it will intersect the Circle on the opposite side, from which Intersecti­ons, Lines drawn into the Center at M, shall be the hours Lines re­quired, to be produced beyond the Center, as many as are needful, which shall be the hours before or after 6 in the Summer half year, the halfs and quarters are after the same manner to be inscribed, by dividing each equal division of the Circle into halfs and quarters.

This Circular Dyalling, was in effect, published and invented by Mr. Foster in his book of a Quadrant, in Anno 1638. for the de­monstration of this Work will demonstrate the truth of those Cir­cular performances, which he Operates on the back of that Qua­drant, but is more expresly hinted in his Posthuma, the demonstrati­on whereof shall follow.

Those that have plain Tables with a Frame, may have Tangent Lines put on the sides of their Frame; and then if a Center be found upon the paper under the Frame by the Intersection of a Ruler laid over those Tangents, the requisite Divisions of a Circle to any Ra­dius that can be described upon the paper, will be most readily given without dividing any Circle or setting of Marks or Letters thereto, the Frame keeping fast that paper, on which the Draught of the Dyal is made, which may also be supplyed from a Circle divided on Pastboard cut out, which is to be Laced upon a Board over the paper whereon the Draught is to be made.

An upright full South Dyal.

This is no other then a Horizontal Dyal, in that Latitude which is equal to the complement of the Latitude of the place, you are in, onely the hours must not be continued beyond the Center, and the Delineation requires no other directions then the former.

On a direct North Dyal, in this Latitude, there will not be above two hours above, and two hours beneath each end of the Horizon­tal Line to be expressed, and the Stile will have the same Elevation and point upward.

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[Page] [Page 9] In all upright North and South Plains, the Meridian or hour line of 12 is perpendicular to the Plains Horizontal line, if the Plain be di­rect the height of the Stile above the substile, is equal to the comple­ment of the Latitude.

If the Plain be South Reclining, or North Inclining, the height of the Stile above the substile, is equal to the Ark of difference be­tween the complement of the Latitude and the Ark of Re-Inclina­tion, and if this latter Ark be greater then the former, the con­trary Pole is elevated.

If the Plain be South Inclining, or North Reclining, the height of the Stile above the substile, is equal to the sum of the Colatitude, and of the Re-Inclination, and when this latter Ark is greater then the Latitude, the Stiles height will be greater then a Quadrant.

Such Horizontal or direct South Dyals, both upright and leaning, whereon the stile hath but small Elevation, are to be drawn with a double Tangent Line without a Center, wherein the following di­rections for direct Polar plains, and those for other Oblique plains, whereon the stile hath but small elevation, will fully direct you.

All Plains that cut the Axis of the World, have a Center; but if they be parallel thereto, the stile hath no Elevation.

Such direct South Reclining, or North Inclining Plains, whose Arch of Reclination is equal to the complement of the Latitude, are parallel to the Axis, and are called Polar plains; in these the Hour Lines will be Tangent Lines of any assumed Radius, and the paral­lel height of the stile above the plain, must be made equal to the Radi­us of the Tangent, by which the hours were set off.

[Page 10]

First draw the Tangent line CG, and perpendicular thereto ACB, upon C as a Center, with any Radius describe the Quadrant of a Circle HB, and prick off the Radius from C to A, from H to E, from B to D, from D to F, from F twice to G.

A Ruler laid from A to E Findes the point 1 on the Tangent CG for the hour line of one, and laid from A to D findes the point 2 for the hour line of two, the hour line of three is at the point H, of four at the point F, of five at the point G; thus are the whole hours easily inserted.

Now to insert the halfs and quarters.

Divide the Arches BE, ED, DH into halfs, and a Ruler laid from A to those divisions wil finde points upon CH, where all the half hours under 45 ^d are to be graduated; and if a Ruler be laid from B to those respective divisions of the Quadrant BH, it will finde points on the Tangent CG, where the whole hours and halfs are to be graduated above 45 ^d, after the same manner are the quarters to be inserted; But in regard the halfs and quarters above 45 ^d will by this direction be found with much uncertainty, I have added the help fol­lowing.

First, Divide the halfs and quarters under 45 ^d as now directed, then for those above make use of this Table.

• [Page 11]In a direct South Polar Dyal.
• The distances of these hour lines.
• Are equal to the di­stances of these hour lines.

• 3 and 3 ¼
• 3 ½ and 4
• 2 ¼ and 3 ¾
• 4: and 4 ¼
• 4 ¼ and 4 ½
• 4 ¼ and 4 ¼
• ¾ and 1 ¼
• 1 ¼ and 2 ½
• 1 ½ and doubled
• ¾ and 1 ¾
• 3 ½ and 3 ¾ doubled: otherwise,
• 1 ¾ and 2 ¾
• 2 and 3 ¾

This I do not assert to be abso­lutely true, but so neer the truth, that there will not arive above one thousandth part of the Radi­us difference in the greatest Dy­al that is made, and will be more certain then any Sector, though of a Vast Radius, or then they can with convenience be prickt down the common way, by a Contingent Line, the meaning of the Table will be illustrated by one or two examples.

The distance between the hour lines of two and a quarter, and three and three quarters will be equal to twice the distance between the hour lines of twelve, and one and a half.

Also the distance between four and a quarter, and four and a half, is equal to twice the distance between three and a half, and three and three quarters; or it is equal to the distance between one and three quarters, and two and three quarters, but the former is nearer the truth.

It will be inconvenient in such plains, as also in direct East or West Dyals, to express any hour line from the substile beyond 75 ^d or 5 hours.

To these I may add some other observations which were commu­nicated by Doctor Richard Sterne, to Mr. Sutton, which may be of use to try the truth of these kinde of Plains.

1. The distance between the hours of 3 and 4, is equal to the di­stance between the hours of 1 and 3.

2. The distance between the hours of 2 and 4, is double to the di­stance between the hours of 12 and 2.

3. The distance between the hours of 11 and 4, is doubte to the distance between 12 and 3, or to that between 9 and 12, and so equal to the distance between 9 and 3, also equal to the distance between 4 and 5.

[Page 12] 4. The distance between the hours of 11 and 5, is double to the distance between the hours of 11 and 4, as also to the distance be­tween the hours of 4 and 5, and between 9 and 3; and qua­druple to the distance between 12 and 9, or between 12 and 3.

These are absolutely true, as may be found by comparing the dif­ferences of the respective Tangents from the natural Tables.

To draw a Polar direct South Dyal.

Having drawn the Plains perpendicular in the middle of the Plain, let that be the hour line of 12, then assuming the stile to be of any convenient parallel height, that will suit the plain, making that Ra­dius, divide a Tangent line into hours and quarters by the former di­rections, and prick them down on the Plain, upon a line drawn per­pendicular to the Meridian or hour line of 12 on each side thereof, and through the points so prickt off, draw lines parallel to the Meridian line, and they shall be the hour lines required, as in the Example.

To describe an Equinoctial Dyal.

Such direct North Recliners, or South Incliners, whose Re-Incli­nation is equal to the Latitude, are parallel to the Equinoctial Cir­cle, and are therefore called Equinoctial Dyals, there is no difficul­ty in describing of these: Divide a circle into 24 equal parts for the whole hours, & afterwards into halfs & quarters, and place the Meri­dian line in the Plains perpendicular, assuming as many of the former hours as the Sun can shine upon for either face, and then placing a round Wyre in the Center for the Stile, perpendicular to the Plain, and the Dyal is finished.

To prick off the Substile and Stiles height on upright Decliners in their true Coast and quantity.

On such Plains draw a Horizontal Line, and cross the same with a perpendicular or Vertical, at the intersection set V at the upper end of the Vertical Line set S, at the lower end N, at the East end of the Horizontal Line set E, at the West end W, prick off the Decli­nation of the Plain in its proper Coast from S or N to D, and draw DV through the Center, the same way count off the Latitude to L, and from it draw a Line into the Center; in the same quarter make a Geometrical square of any proportion at pleasure, so that two sides thereof may be parallel to the Horizontal and Vertical Line, at the intersection of one of the sides thereof, with the Vertical set A, and of the other side, with the Horizontal Line, set B, and where the Latitude Line intersects the side of the square, set F.

To prick off the Substilar.

For Latitudes above 45 ^d take BF, and prick it on the Line of De­clination beyond the Center from V to O, and from the point O, draw the Line OC parallel to the Vertical Line, and produced be­yond O, if need require, and thereon from C to I prick the side of the Square, and a line drawn into the Center, shall be the Substilar line; but for Latitudes under 45 ^d prick the side of the square from V to O, and draw OC as before, and make CI equal to AF, and a line from I drawn into the Center, shall be the Substilar line.

Stiles Height.

From the Point I erect the extent OC perpendicularly to the sub­stilar OI, at the extremity thereof set K, from whence draw a line into the Center, and the Angle IVK will shew how much the stile is to be elevated above the substilar line.

In order to the Demonstration hereof, let it be observed that an Angle may be prickt off by Sines or Tangents in stead of Chords.

To prick off an Angle by Sines or Tangents.

As in the Scheme an­nexed, let BC be Radius, and let there be an arch prickt off with Chords, as BE; I say, if the Tan­gent of the said Ark BA be taken out of a line of Tangents to the same Radius, and be erected perpendicular to the end of the Radius, as BA, a line drawn from A into the center, shall include the same An­gle as was prickt off by Chords, as is evident from the definition of a Tangent.

In like manner an Angle may be prickt off by Sines, the nearest di­stance from E to BC is the Sine of the Arch BE, so in like manner the nearest distance from B to EC, is the Sine of the same Arch; wherefore if with the Sine of an Arch from the end of the Radius be described another Ark, as F, and from the Cen­ter or other extremitie of the Radius, a line drawn just tou­ching the same, the Angle included between the said line, and the Radius shall be an Ark equal to the Ark belonging to the said Sine; and what is here done by a line of natural Sines or Tangents, by help of a Decimal line of equal parts, equal to the Radius' may be done by help of the natural Tables without them.

Any Proportion relating to the sixteen cases of Sphoerical Triangles, amongst which the Radius is always ingredient, may be so varyed, that the Radius may be in the third place, and a Tangent or a Sine in the fourth place, and then if the Ark belonging to the fourth Proportional be known, an Angle equal thereto may be prickt off by Tangents or Sines according to the nature of the fourth term, as be­forth; if it be unknown, notwithstanding an Angle equal thereto may be prickt off with Sines or Tangents according to the nature of the fourth term, from the two first terms of the Proportion, because they are in such Proportion, the first to the second, as the Radius or third term is to the fourth; upon this Basis follows the demonstra­tion of this Scheme.

The Demonstration of the former Scheme for upright Decliners.

To draw the hour Lines of the former upright Decliner.

First draw the Plains perpendicular CO, which for upright Plains [Page 17] is the Meridian line, and with the Radius VF of the former Scheme, draw an occult Ark upon C as a center, and therein set off NV equal to TN of the former Scheme, and draw the line IVC for the substile, and with the Radius VC describe a circle thereon.

1. The first work will be to finde the Regulating Point in the Substile, called the Point Sol.

Prick the Radius of the said Circle from I to P, and from C to Q, and draw the Line IP, and therein make IK equal to IK in the former Scheme, then a ruler said from K to Q, where it intersects the Substilar is the Regulating Point ☉.

2. To finde a Point from whence the Circle is to be divided into 12 equal parts.

Lay a Ruler from O to the point ☉, and it intersects the Circle on the Opposite side, at it set M.

3. To divide the Circle.

From the point M before found, lay a ruler through the Center V, and it will finde a Point on the other side the circle, at it set f, the points Mf divide the Circle into halfs, and each half is to be divided into six parts, the Radius VM will easily divide each half into three parts, and then it will be easie to divide each of those parts into halfs, and so the whole Circle will be divided into 12 equal parts, and if it be desired to inscribe the half hours and quarters, then must each of those parts be divided into halfs and quarters.

Having thus divided the Circle into 12 parts, distinguish them with the Letters of the Alphabet.

4. To draw the hour lines, and to number them.

Lay a Ruler to each of those Letter divisions, and upon the Regu­lating Point ☉ in the Substile, and it will finde points on the Oppo­site side of the Circle from which if lines be drawn into the Center at C' they shall be the hour lines required.

5. An inconvenience shunned.

When the points through which the hour lines are to be drawn, fall near the Center, the hour lines cannot be drawn from the said points with certainty.

[Page 18] In this case let any hour line near the said point be produced, if need require, and upon C as a Center with the Radius CV describe the Ark of a Circle from the produced hour-line, take half the di­stance between the point, where the produced hour line, cuts the Circle, and the point through which the hour line proposed is to be drawn, and prick that extent off in the Arch swept, setting one foot where the Arch swept cuts the hour line already drawn, from whence half the distance was measured, and the other foot will finde a Point in the said Arch, through which the hour line desi­red is to pass: Thus we inscribed the hour line of three in the former Dyal.

When the North Pole is elevated, the Center of the Dyal must be below, but when the South Pole is elevated, as in this example, it must be above, and the Point ☉ must be alwayes found in the other part of the Diameter most remote from the Center.

6. The Stiles height may be transferred from the former Scheme into this, by pricking the Arch TK twice in this Circle, and it findes a point, from whence a line drawn into the Center, shall re­present the Stile.

That the hour lines are true Delineated.

The Point Sol found in the Substile, divides the Diameter of the Circle drawn on the said Substile in such Proportion, as the Radius is to the Sineof the Stiles height, by reason of the equiangled Tri­angles, whose Bases bear such proportion as their perpendiculars.

The next work performed by dividing the Circle into 12 equal parts, and finding points on the opposite side, by laying a ruler over those equal parts through the point Sol, carrieth on this Proportion.

In the third figure following, let BA represent the Substile, and let the Regulating Point be at M, so that BM bears such Proportion to MA, as the Radius doth to the Sine of the Stiles height, let the perpendicular BHE represent a line of Tangents, whose Radius is equal to BC, and the perpendicular AD, another line of Tangents [Page 19] to the same Radius both infinitely produced, and let BE represent the Tangent of some particular Arch or hour to be drawn, from E through the point M draw a right line to D, and the Proportion lies evident in right lines.

If from every degree of a Tangent line, lines be drawn into the Center of the Circle proper to that Tangent, they shall divide that Quadrant to which they belong into 90 equal parts, this follows from the definition of Tangents, but if the lines be drawn from every degree of the said tangent to the extreamity of the diamer, they shall divide a Semicircle into 90 equal parts by 20 Prop. 3. Euclid. because an Angle in the Circumference, is but half so much as it is in the Center.

Thus the Semicircle BGA is supposed to be divided into 90 equal parts from the Tangent BE, by lines drawn to A, as is the line EGA, so is the other Semicircle from the other Tangent AD, by lines drawn to B, as is BFD; Now if it can be demonstrated that the points GMF are in a right line, it follows that the same Proportions may be carried on from the equal divisions of these Semicircles, as were done in the right Tangent lines BE and AD, this Proposition being very well Geometrically demonstrated by my loving friend Mr. Thomas Harvy; his Demonstration thereof shall hereafter follow.

Having found the distance of the hour line, from the subsistle AB, if B be made the Center of the Dyal, the right line BF drawn into the Center, shall represent the hour line proposed, making the same Angle with the Substile, as was found by the proportion; note that no hour can be further distant on one side or other of the Substile then 90 ^d, and the said Ark of distance will be found in one of the Semicircles, and if the Center were not placed in the Circumfe­rence, the Angle found would be extended beyond its due quantity, in all upright Decliners or leaning Plains: The first Proportion carried on, is this;

Whereby is found the point from whence the Circle is to begin to be divided into 12 parts for the whole hours with their halfs and [Page 20] quarters, and those equal divisions give the Angles between the Me­ridian of the plain, and the respective hours, called by some the An­gles at the Pole, and then the work is the same as before: Now to the Proposition.

Construction.

The right lines AD, BE are parallel, and touch the Circle in the extreams of the Diameter AB, DE is drawn at pleasure, cutting the Tangents in D and E, and the Diameter in M, BD and AE cut­teth the Circumference in F and G, it is required to be proved that the Points F, M, and G, are in a right line.

Draw the lines MF, MG, first if AD be = BE then AM is = MB, for the Triangles ADM and BEM are equiangeld, therefore M is the Center. Again AD being = BE and the Angle DAB = EBA & AB common to both Triangles DAB, EBA therefore the Angles DBA, EAB are equal, and the double of them FMA, GMB the Angles at the Center, are also equal, wherefore FM and MG is one and the same right line which was to be proved.

Secondly, if AD be not equal to BE, as in the second, third, & fourth Scheams let BE be greater, and make BH = AD and BI = AM, and draw the right line AH cutting the Circumference in K; draw likewise IK, IH, GK, and extend GK infinitely both wayes, which shall be either parallel to AB (as in the second figure) or cut it in the point P one side or the other, as in the third and fourth fi­gures, in each from the points A and B, and the Center C at right Angles to GK, draw the lines AO, BN, CL lastly draw BG, from which Construction it willfollow, That

• 1. HI is parallel to ED
  • for HB is = AD and BI = AM and the Angles HBI and DAM are equal, therefore the angle AMD = BME is = BIH wherefore HI is parallel to ED.
• 2. IK is parallel to MF
  • for HB is = AD and the Angle HBA = DAB, and AB common to both the Triangles HBA and DAB, therefore the Angles HAB and DBA are equal, wherefore AK = BF, but AK being = BF and IA = MB, and the Angle KAI = MBF, therefore KI is parallel to MF.

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[Page] [Page 21] 3. The Triangles AHE and AGK, are equiangled, so also are the Triangles AHB and AGO, because the Angle AKG = ABG is = AEB and EAH common to both the Triangles AHE and AGK, therefore they are equiangled; again the Angles AHB and and AGO being equal, the right Angled Triangles ABH and AOG are also equiangled.

4. NL is equal to LO, because BC is equal to CA and NK is = GO, because KL is = LG; now in the equiangled Triangles ABH, AOG as BH: HA ∷ OG: GA and in the Triangles AHE, AGK as HA: HE ∷ GA to GK, therefore (ex equo) it will be as BH: HE ∷ OG: GK, but as BH: HE ∷ BI: IM for IH is parallel to EM, therefore as BI: IM ∷ OG: GK but NK is = OG, therefore BI: IM ∷ NK: KG, but IC is half of IM and KL half of KG, therefore BI: IC ∷ NK: KL, therefore by Compo­sition of Proportion, as BC: IC ∷ NL: KL, alternately, as BC: NL ∷ IC: KL.

In the second figure BC is = NL, therefore IC is = KL, where­fore (because IC is also parallel to KL) IK is parallel to CL, by like reason GM is also parallel to CL, but in the third and fourth figures the sides PC, PL of the Triangle PCL are cut Proportionally by the parallel BN, and as before, as BC: NL ∷ IC: KL, and as PC: PL ∷ BC: NL, therefore PC: PL ∷ IC: KL, wherefore IK is parallel to CL.

Again as before, BC: NL ∷ IC: KL, But CM is = IC and IG is = KL, therefore as BC: NL ∷ CM: LG, but as hath been said, as BC: NL ∷ PC: PL, therefore as PC: PL ∷ CM: LG, where­fore because the sides PM: PG, of the Triangle PMG are cut proportionally in C and L, CL and MG are parallel one to ano­ther.

Now in all the three figures it hath been proved that IK and MG are either of them parallel to CL, therefore they are parallel one to another, wherefore the Angle GMI is equal to the An­gle KIB which before was proved to be equal to AMF, there­fore the Angles GMI and AMF are equal one to another, where­fore FM and MG is one and the same right line which was to be proved.

To draw an East or West Dyal.

Let the hour line of six, which is also the Substilar line, make an An­gle equal to the Latitude of the place, with the Plains Horizontal line, above that end of it that points to the Coast of the elevated Pole, then draw a line perpendicular to the Substilar line, which some call a Contingent line, and with such a Radius as you determine the Stile shall have parallel height above the Substile: divide a Tangent line of hours and quarters according to the direction for direct Polar Recliners, then through those divisions draw lines parallel to the Substile and they shall be the hour lines required; thus the hours of 5 and 7, are each of them Tangents of 15 ^d from six, and so for the rest, let them be numbred on each side of six (being the Substile) for an East dyal with the morning hours, for a West dyal with the afternoon hours, the one being the complement of the other to 12 hours, and therefore we have but one example, namely an East dyal for the Latitude of London.

How to fill this or any other Plain, with any determined number of hours, shall afterwards be handled.

To prick off the requisites of an East or West Reclining or Inclining Dyal in their true Scituation and quantity.

In these Plains, first draw the Plains perpendicular, and cross it with a Horizontal line, which is also the Meridian line; in any one of the quarters, make a Geometrical square, as before directed for up­right decliners, and from the Plains Vertical in the said quarter, count off the Latitude to L, and from the Horizontal line in the said quarter count off the Reclination or Inclination to R, and from L and R, draw lines into the Center, and where the Latitude line intersects the side of the square, set F.

1. To prick off the Substile.

For Latitudes above 45 ^d place BF from V the Center in the Hori­zontal line for Recliners Northwards, for Incliners Southwards, and thereto set O, and through the said point, draw a line parallel to the Vertical, and place the nearest distance from A to RV on the former parallel, from O for Recliners upwards to I, but for Incliners downwards, and a line thence drawn into the Center, shall be the Substile.

2. The Stiles height.

Place the nearest distance from B to RV on a perdendicular raised from the point I in the Substile, and it findes the Point K, whence a line drawn into the Center, shall represent the Stile. In the other Hemisphere the words Northwards and Southwards, must be mu­tually changed.

For Latitudes under 45′

The side of the square must be placed from V to O and AF must be placed on the line of Reclination or Inclination from V to C, the nearest distance from C to VA placed on the perpendicular, passing through O for Recliners upwards, but for Incliners down­wards, findes the Point I, through which the Substile is to pass, and the nearest distance from C to VB raised perpendicularly on the point I in the Substile, findes the point K for the Stile, as before.

Demonstration. 1. For the Substile.

In these Plains VB or VA being Radius FB is the Contangent of [Page 24] the Latitude and the nearest distance from A to RV, is the Cosine of Reclination or Inclination to the same Radius, and the nearest di­stance from B to RV, the Sine thereof; this for Latitudes above 45, but for lesser Latitudes. AF the Tangent of the Latitude was made Radius, and thereupon the Radius or side of the square be-became the Cotangent of the Latitude, and the nearest distance from C to VA was the Cosine, and from C to VB the Sine of the Reclination or Inclination, so that the prescribed Construction in both cases erects the Cosine of the Re-Inclination on the Cotangent of the Latitude, which is made good from this proportion.

2. For the Stile.

A proportion that will serve to prick it off, is,

Admit the Stiles height were calculated, and truely prickt off by Chords, I say if the Sine of the Latitude be made a Radius, as the side of the square is in Latitudes above 45 ^d and as AF is in Lati­tudes under 45 ^d the Radius to the said Sine will become the Cose­cant of the Latitude; now we shall prove that VK is the Cosecant of the Latitude, the three squares IK, IO, and OV, are equal to the square of VK, as before asserted, two of these squares IK and IO are equal to the square of the Radius, the third square, viz. of VO is the square of the Cotangent of the Latitude, but the square of the Radius, more the square of the Cotangent of the Latitude is equal to the square of the Cosecant of the Latitude; I say then, if from K a perpendicular be let fall to VI, so as that it may be equal to the Cosine of the Re-Inclination shall prick off the Stile true by the directions for pricking off an Angle by Sines, whereto the proportion for the Stiles height is suited and would pass through the point I, by

[Page 25] reason of the equality of the three squares to the square of VK, then by consequence the contrary follows, having the point I first given, the Point K may be found according to the prescribed Constru­ction.

The Point Sol, Substile, Stile, and hour-lines, are all found after the same manner as in an upright Decliner, the Substile being set off from the Meridian line here, after the same manner as it was from the Meridian there.

To prick off the Requisites in all Declining, Reclining, and Incli­ing Plains in their true Coast and quantity.

UPon any Plain first draw a true Horizontal line, at the East end thereof set E, and at the West end W, cross the said line with a perpendicular, which may be called the Plains perpendicular, by some termed the Plains Vertical line, or line of Reclination, at the intersection of these two lines set V, and upon it as a Center de­scribe a Circle (the Radius whereof may be equal to 60 ^d of a line of Chords) at the upper end of the Vertical line set S, and at the lower end N.

As the Declination is, set it off with Chords from S or N towards the true Coast, and at it set D, from whence draw a line through the Center; also set off the Latitude of the place the same way and in the same quarter, and at it set L, from which draw a line to the Center.

From that end of the Horizontal line, towards which the declina­tion was counted, set the Inclination (which as well as the Reclina­is reckoned from the Zenith, the former being the Denomination of the under, the latter of the upper face) upwards towards S, and the Reclination downwards towards N, and at it set R, from whence draw a line through the Center to the other side, of the Circle.

[Page 26] In the same quarter of Declination, draw HA parallel to the Hori­zontal line, and FG parallel to the Vertical line, in a Geometrical square, of like and of any convenient distance from the Center at pleasure, and where the Latitude line intersects the side of the square, let the letter F be placed.

On the line of Declination beyond the Center, make VO equal to FG, and draw OB parallel to the Horizontal line, continued till it meet with the side of the square FG produced, at the point of con­currence, set B, and where it intersects the Plains perpendicular, set P, and draw OC parallel to the Vertical line cutting WE at C, and make HA equal to OC or BG, now by help of the three points A, B, C thus found, the requisites will be easily prickt off.

Annotations on the former Scheam.

1. That for Latitudes under 45 ^d this construction of the Scheam, supposeth the sides of the square produced, which will therefore be lyable to large excursions or other inconveniences, wherefore for such Latitudes, I shall somewhat vary from the construction pre­scribed.

2. In finding the Substilar line, in stead of erecting CI upon VC, you may prick the same on the Vertical line VN, and thereto erect VC, and get the point I possibly with more certainty by finding the intersection of two Arks where the said Point is to pass.

3. In pricking off the Meridian line, the distance of C from the Center may be doubled or tripled, but so must likewise VP, and the nearest distance from C to RV erected on a line drawn parallel to WE, passing through the Point P so found, and in stead of drawing such a parallel, the Point M may be found by the intersection of two Arks.

4. That this Scheam placeth the requisites of all Dyals in their true coast and quantity, yet notwithstanding if this Scheam be held before a Looking-Glass, the Effigies thereof in the Glass shews how the Scheam would happen and place the Requisites, namely, the Stile, Substile and Meridian, for a Plain of the same Denomination, but declining to the contrary Coast.

And if the face of the said Scheam be laid upon a Window, and the Substile, Stile and Meridian be continued through the Center on the backside thereof, it shews you how these requisites are to be pla­ced on the opposite side of the Plain, which being done, may be held before a Looking-Glass as before, and will be represented for the contrary Declination of that opposite face: the truth of all which will be confirmed from the Scheam it self.

[Page 28] This Scheam for Declining, Reclining, or Inclining Plains, useth a new method of Calculation, derived from an Oblique Triangle in the Sphere, wherein there is two sides with the Angle comprehended, given to finde both the other Angles, which is reduced by a perpen­dicular to two right Angled Triangles, from which the following proportions are derived.

I shall therefore first deliver the said method, then demonstrate that the said proportions are carried on in the Scheam; and lastly, from the Sphere, shew how those proportions do arise.

Affections of a Polar Plain.

The Substilar on the upper face, lies above that end of the Hori­zontal line, towards the Coast of Declination, and the Meridian lyes parallel to the Substile beyond it, towards that end of the Hori­zontal line that is towards the Coast of Declination.

For Declining, Reclining, or Inclining Plains.

First finde a Polar Plains Reclination for the same Declination.

Then for South Recliners and North Incliners, get the difference, but for North Recliners and South Incliners, the sum of a Polar plains Reclination, and of the Re Inclination of the plain proposed, and then it holds.

Directions for the true Scituating of the Meridian and Substile suited to the former method of Calculation.

To place the Meridian.

On all North Recliners the Meridian lies below the Horizontal Line from that end thereof, opposite to the Coast of Declination, be­cause at noon the Sun being South, casts the shadow of the Stile to the Northwards.

On the under face, being a South Incliner, it must always be placed below the Horizontal Line, below that end of it toward the Coast of Declination.

These Directions suppose the Declination to be denominated from the Scituation of that face of the plain on which the Dyal is to be made, and the Horizontal line for all Dyals that have Centers, is supposed to pass through the same.

Now to the Demonstration of the former Scheam.

1. 'Tis asserted that if RV fall into the point B, the plain is a Polar plain, in which case the Stile is parallel to the Axis of the world.

Demonstration.

Every Declining plain may have such a Reclination found thereto, as shall make the said plain become a Polar plain, and the propor­tion to finde it, may be thus:

In the former Scheam. if we make FG the Cotangent of the La­titude [Page 32] Radius, the side of the square will be the Tangent of the Lati­tude, now VO equal to FG, being Radius, OC equal to GB, is the Cosine of the Declination; wherefore a Line drawn into the Center from B, shall include the Angle of a Polar plains Reclination agree­able to the two first terms of the proportion, and to the directions for pricking off an Angle by Tangents.

2. That the substile is true prickt off.

Upon V as a Center with the Radius VB, imagine or describe a Circle then is BG equal to VP, the Sine of a polar plains Reclinati­on, which is equal to HA, and the Ark comprehended between A and B, will be a Quadrant.

But in a Quadrant any line being drawn from the Limbe passing through the Center, the nearest distance from the end of one of the Radij will be the Sine of the Ark thence counted, and the nearest distance from the other Radius thence counted, will be the Sine of the former Arks complement; so in this Scheam the nearest distance from B to RV, when a plain Reclines, is the Sine of the Ark of dif­ference, but when it inclines of the sum of the Reclination of the Plain proposed, and of the Reclination of a Polar plain, and the nearest distance from A to RV, is the Cosine of the said Ark.

Make VQ on the plains perpendicular equal to IC, which is equal to the nearest distance from A to RV, and from the point Q erect the perpendicular QT, whereto the line PO will be parallel, and con­sequently there will be a proportion wrought: Thus it lyes,

These two terms are of one Radius, namely VB,

These two terms are to another Radius, namely QV, then if the Tangent of an Ark be erected on its own Radius, as here is QI equal to CV, and a line be drawn from the extreamity into the Cen­ter, the Angle belonging to that Tangent shall be prickt off agree­able to the general direction.

3. That the stiles height is true prickt off.

The Proportion altered to bring the Radius in the third place will be,

In the Scheam making VQ Radius, VI becomes the Secant of the substiles distance from the Plains perpendicular, and the nearstdie­stance from B to RV, is the tangent of the difference or sum of the Reclinations, which when VB was Radius, was but the Sine there­of, the reason why it now becomes a tangent, is because the Consine of the said Ark VQ, is made Radius: But,

Therefore the nearest distance from B to RV equal to IK, being erected thereon, and a line from the extreamity drawn into the Center, shall prick off the stiles height suitable to the two first terms of the former proportion, and to the general direction for pricking off an Angle by Tangents.

4. That the Meridian is true prickt off, the Proportion to effect it: Is,

If VC be made Radius, then is CO equal to VP the Tangent of the complement of the Plains Declination, and the nearest distance from C to RV, is the Sine of the Re Inclination to the same Radius, which is erected perpendicularly on VP suitable to the two first terms of the proportion and the General Direction.

Lastly, that VK is equal to VF, these Symbols are used, q signi­fieth square, + for more or Addition, = equal.

VBq = VQq + IKq, the reason is because VB being Radius, VQ is the Cosine of an Ark to that Radius, and IK the Sine by con­struction.

[Page 34] VBq = VGq + GBq, therefore these two squares are equal to the two before.

If to the latter part of each of these Equations, we adde QIq or rather its equal POq, the sum shall be equal to VKq.

I say then VQq + IKq + POq = VKq, this will be granted from the former Demonstration for upright Decliners.

Again,

VGq + GBq + POq = VFq, therefore VF is equal to VK, this cannot be denyed, because, VGq + FGq = VFq.

And the two squares GBq, or rather VPq + POq are equal to GFq, which is equal to VOq by Construction.

To draw the hour-lines for the former South Plain, Declin­ing Eastwards 40 degrees, Inclining 15 degrees Lati­tude, 51 degrees, 32 minutes.

First having assigned the Center of the Dyal, through the same draw the plains perpendicular, represented by the prickt line CN, and with the Radius of the former Scheam upon C as a Center, describe an Occult Ark, and therein set off NV equal to the substiles distance from the plains perpendicular, and through the point V draw the Substilar, and upon V as a Center, describe the Circle, and prick off the Meridians distance from the the Substile in the former Scheam, namely YX twice in this Circle from I to O, and draw CO for the Meridian, after the same manner set off the Stile, then finde the Regulating point Sol, divide the Circle, and draw the hour-lines ac­cording to former Directions, and when hour-lines are to lye both above and below the Center, they are to be drawn through.

To fit the Dyalling Scheam, for Latitudes under 45 degrees. A South Plain Declining 50 degrees East, Inclining 20 de­grees, Latitude 30 degrees.

The former Construction would serve, if the sides of the square were produced far enough, but to shun any such excursion, make VO equal to the side of the square, and through the point F, draw a parallel to the Plains perpendicular, and where the parallel OP pro­duced interesects it, is the point B, upon V as a Center, with the ex­tent

[Page 35] VB, draw the Arch ZB, and prick off a Quadrant thereof from the point B, and it will finde the point A, the point C is found no otherwise then before; and now having these three points, the whole work is to be finished according to those Directions, to which when the Stile hath a competent height, nothing need more be add­ed, unless it be some examples.

How to draw such Dyals whereon the Stile hath no elevation as Polar Plains, or but very small elevation, as in up­right far Decliners, and many leaning Plains.

These plains are known easily, for if the Re-Inclination pass through the point B, the plain is a Polar plain, and the Substile is to be found by the former Construction, which the Scheam makes the same with the Meridian: moreover, another Ark is to be found, called the In­clination of Meridians: The proportion to finde it, is:

If CO be made Radius, CV will be the Tangent of the Declina­tion, which enter on VL from V to K, and the nearest distance from K to SV shall be the Tangent of the Inclination of Meridians, which is to be prickt on its own Radius from P to M, and draw a line from the Center, passing through M to the Limbe, whereto set F, so is the Arch NF the Inclination of Meridians sought, to wit, the Arch of time between the substile and Meridian.

To draw the hour-lines.

First draw a perpendicular on the Plain VN, and upon V as a Center, describe the Arch of a Circle, and from the Dyalling Scheam prick off the substiles distance, and draw IV which shall represent the same, as also the Inclination of Meridians from I to f, then upon V as a Center, describe as great a Circle as the plain will admit, and finde the point f, therein also, by laying a Ruler over V the Center, and f in the former Circle, and from the said point divide the Circle into twenty four equal parts for the whole hours (but we shall not need above [Page 36] half of them) then determine what shall be the parallel height of the Stile above the Substile, and prick the same on the Substilar line, from V the Center to I, through which point draw a Contingent line at right Angles to the Substile, and laying a Ruler from the Center over each of the divisions of the Circle, through the points where it intersects the Contingent line, if lines be drawn parallel to the Substile, they shall be the hour-lines required, the hour-line that belongs to the point f being the Meridian-line or hour-line of 12, af­ter the same manner are the halfs and quarters to be inscribed, if the Contingent line be too high, the Center V may be placed lower, if it be required, to fit so many hours precise to the Plain; first draw it very large upon some Floor, and then it may be proportioned out for a lesser Plain at pleasure, as was mentioned for East or West Plains.

A South Plain, Declining Eastwards, 30 degrees, Reclin­ing 34 degrees, 31 minutes.

[Page 37] The Scheam placeth all things right for Equinoctial Reclining Plains, without any further caution.

A South Plain Declining 30 degrees Eastwards, Reclining 25 degrees, Latitude 51 degrees 32 minutes.

In these Plains, because the hour-lines will run close together, the Dyal must be drawn without a Center, by help of two Contingent lines, and first of all the Inclination of Meridians must be known, thereby is meant the Arch of time between the Substile and the Me­ridian line or hour line of 12, and that may be found several ways; here I shall follow the proportion in Sines before delivered.

Making VI Radius prick the same on the Latitude line VL, and from the point found, take the nearest distance to the Horizontal line, place this extent from V to F, and draw CF; then take the nearest distance from R to the Plains Vertical SN, which place from V to Q, then draw QT parallel to FC, so is VT the Sine of the Inclination of Meridians, which may be easily measured in the Limbe by the Arch SY, by drawing a line parallel to the Plains perpendicu­lar from the point T, or by pricking the same on HA produced, and laying a Ruler thereto, or by drawing the touch of an Arch with VT upon S as a Center, then a Ruler laid from V touching the outward Extremity of that Arch, findes the point Y.

Moreover, we need not make VI Radius, but prick the nearest di­stance from L to VE, from V upwards, then the nearest distance to the Plains perpendicular from the Intersection of the Substile, with the Limbe must be placed from V towards W, and from the two points thus found a line drawn, and the rest of the work, as before.

To draw the Hour-lines.

First draw the Plains perpendicular CN, and draw an Occult Arch, wherein prick down NY and NK, and draw the Substile and Stile as before, making the same Angles.

Through any two points in the Substile, as at A and B, draw two right lines continued, making right Angles therewith.

Draw a line parallel to the Stile at any convenient distance, which is to represent the new Stile, as here DE.

[Page 38] Take the nearest distance from B to DE, and set it on the Substile from B to V, also the nearest distance from A to DE, and set it from V to C, through which point draw another line perpendicu­lar to the Substile.

Upon V as a Center, describe the Arch of a Circle of as large a Radius as the plain will admit, and from the substile on the same side thereof the Meridian happened in the former Scheam, set off the In­clination of Meridians, and it findes the point M, from whence di­vide the Circle into 24 equal parts, and draw lines from the Cen­ter V through those parts, cutting both the Contingent lines B and C, the respective divisions of the Contingent line C, must be transferred into the Contingent line A, and there be made of the like distance from the substile as in the said line C, then lines drawn through the Divisions of the two Contingent lines A and B, shall be the respective hour-lines required.

A South Plain Declining Eastwards 30 degrees, Reclining 25 degrees, Latitude 51 degrees, 32 minutes.

[Page 39] After the same manner must such East and West Recliners or In­cliners, that have small elevation of the Stile, and upright far decli­ners be pricked down, and in these plains the Meridian many times must be left out, the proportion to finde the Inclination of Meridi­ans for upright Decliners: Is,

In the Dyalling Scheam, making CV Radius, CO is the Cotan­gent of the Declination, which enter on the Latitude line VL, and take the nearest distance to SV, which extent prickt upon CO, and it findes a Point, through which a line drawn from the Center to the Limbe, shall shew the inclination of Meridians to be measured from N.

Otherwise:

That we may not transfer large Divisions on the Contingent line from a small Circle, and that the Plain may be filled with any deter­mined number of hours, such as by after Directions shall be found meet, draw any right line on a Board or Floor that shall represent the Plains perpendicular, as CN, and from the same set off the Sub­stile and Stiles height from the general Scheam as before, drawing a line parallel to the Stile as DE, also a line perpendicular to the Sub­stile, which I call the Floor Contingent, and that it may be large, let it be of a good distance from the Center; from the point of Inter­section at Y, take the nearest distance to the parallel Stile, which prick from Y to V, and upon V as a Center describe as large a Circle as may be with convenience, and from the Substile set off the Incli­nation of Meridians therein to M (which Ark refers to V as its Cen­ter) and from the said Point divide it into 24 equal hours (or fewer, no more then are required) and laying a Ruler over V, and those respective divisions graduate them on the large Contingent line, I say from this large Contingent line thus drawn and divided, we may proportion out the Divisions of two (or many) Contingent lines that are lesser, and thereby fill the Plain with any proper number of hours required.

Then in Order to drawing the hour-lines on the Plain.

[Page 40] The first work will be to draw the larger contingent line on the plain, which may be drawn any where at pleasure: for performing where­of, note, that what Angle the Substile makes with the Plains perpen­dicular, the Contingent line is to make the same with the Horizon­tal line, and the complement thereof, with the Vertical line; also draw another Contingent line above this, parallel thereto at any convenient distance.

In the bigger Contingent line assume any two points to limit the outward most hours that are intended to be drawn on the Plain, as admit on the former Plain, I would bring on hours from six in the morning, to two in the afternoon, between the space A and B of the greater Contingent, take the said extent AB, and upon the point B at 2 of the Floor Contingent, describe an Ark therewith, to wit, L, then from A Draw the line AL, just touching the outward extrea­mity of the said Ark.

I say the nearest distance from D to AL, being pricked on the Plains greater Contingent from A to D, findes a point therein through which the Stile is to pass.

Also the nearest distance from Y to AL, findes the space AY on the Plains greater Contingent, and through the point Y a line drawn perpendicular to AB, shall be the Substilar line.

The Stile is to be drawn through the Point D, making an Angle with the Plains Contingent line equal to the complement of its height above the Substile in this example 80 ^d 57′, to wit, the Arch NY.

The respective nearest distances to AL, from each hour-point, in the Floor Contingent, being pricked on the Plains greater Contin­gent from A towards B, findes Points therein, through which the hour-lines are to pass.

The next work will be to limit one of the extream hours on the Plains lesser Contingent, and that must be done by proportion.

[Page 41] This proportion is to be carryed on in the Draught on the Floor, place DY from A to G on the Floor Contingent, and with the ex­tent EG taken from the Plains lesser Contingent, upon G on the Floor Contingent, draw the Arch O, and from A draw a line touch­ing the outward extreamity of the said Arch, and let it be produced.

The nearest distance from Y to AO, being prickt on the Plains lesser Contingent, reaches from G to C, the point limiting the outward hour of six.

Then if the nearest distances to the line AO, be taken from all the respective hours on the Floor Contingent, and placed on the Plains lesser Contingent from C towards F, you will finde all the hours points required, through which and the like points on the Plains greater Contingent, the hour-lines are to be drawn.

Here note, that the extent DY on the Floor Contingent, may be doubled or tripled; if it be tripled it reacheth to H, also the extent EG on the plains lesser Contingent, is to be encreased after the same manner, and an Ark therewith described on H before found, as Q, and by this means the line AO will be drawn and produced with more certainty, then by the Ark O near the Center.

And what is here done by help of the Stiles distance from the Substile, may be done by help of the outward hour, if the distance of the said hour-line from the Substile be found Geometrically or by Cal­culation, for the said hour-line will make an Angle with the Plains Contingent lines, equal to the complement of the Ark of its distance from the Substile.

This Plain is capable of more hours which cannot conveniently be brought on.

After the same manner are upright far Decliners to be dealt with­all, and all other Plains having small height of Stile.

But to limit the outward hours on Polar plains, and East or West Plains, the trouble will not be half so much.

A South Plain Declining Eastwards 30 degrees, Reclining 25 degrees, Latitude 51 degrees, 32 minutes. A second method of Calculation for Oblique Plains.

By the former method the Meridians distance from the Plains per­pendicular is to be found, and the Polar Reclination Calculated.

[Page 42] Then for South Recliners, or North Incliners, get the difference, but for North Recliners or South Incliners the sum, of the Polar Plains Reclination, and of the Reclination of the Plain proposed, and it holds.

Which Pole is elevated is elevated is easily determined, by comparing the Reclination of the proposed Plain with the Polar Reclination, and all other affections are to be determined, as in the first Method.

Then for the Substile, and Inclination of Meridians.

This method ariseth from the aforementioned Oblique Triangle in the Sphere, in which by help of two sides, and the Angle compre­hended, the third side is first found, and the other Requisites by the proportions for Opposite sides and Angles.

Proportions for upright Decliners. 1. To finde the Substiles distance from the Meridian.

2. Angle of 12, and 6.

3. Inclination of Meridians.

4. Stiles height.

These Arks are largely defined in my Treatise, The Sector on a Quadrant.

The complement of the Latitude of the place, is such a new La­titude: in which they shall stand as upright Plains, and the comple­ment of their Re-Inclination is their new Declination in that new Latitude, having thus made them upright Decliners, the former pro­portions will serve to Calculate all the Requisites.

In all upright Plains, the Meridian lyeth in the plains perpendicu­lar, and if they Decline from the South (in this Hemisphere) it is to descend or run downward; if from the North it ascends, and the Substile lyeth on that side thereof opposite to the Coast of Decli­nation.

In East or West Re-Incliners, it lyeth in the plains horizontal line, on the Inclining side the South Pole is elevated, but on the upper side the North Pole, and the Substile lyeth above or below that end of the Meridian line, which points to the Pole elevated above the Plain.

On all plains whatsoever to Calculate the hour distances.

By the Angle at the Pole, is meant the Ark of difference between the Ark called the Inclination of Meridians, and the distance of any hour from the Meridian for all hours on the same side of the Meri­dian the Substile falls, and the sum of these two Arks for all hours on the other side the Meridian.

[Page 44] All hours on any Plain go to the contrary Coast of their Scitua­tion in the Sphere, thus all the morning or Eastern hours, go to the Western Coast of the plain, and all the evening or Western hours, go to the Eastern Coast of the Plain.

They may be referred to a new Latitude, in which they shall stand as upright Plains, and then they will have a new Declination in that new Latitude; which two things being found, the former Proporti­ons for upright Decliners will serve to Calculate all the Arks re­quired.

How this may be done on a Globe, is not difficult to apprehend, having set the Globe to your Latitude, let one of the Meridians of the Ecliptick or Longitude in the heavens, represent a Declining Re­clining Plain, this Circle intersects the Meridian of the place in two Points, the one above, the other beneath the Horizon: Imagine the Globe to be so fixed, that it cannot move upon its Poles, then ele­vate or depress the Globe so in the Meridian that the point of Inter­section above the Horizon may come under the Zenith, then will the Pole of the world be elevated above the Horizon to the new Lati­tude sought, and where the Meridian of Longitude that represents the Plain intersects the Horizon it shews the new Declination.

Or it may be thus apprehended: The distance between the Pole of the world, and that point of Intersection that represents the Zenith of the new Latitude, is the complement of the said new Latitude, and the distance between that point, and the Equinoctial is the new Latitude it self; the new Declination is the complement of the Angle between the plain and the meridian of the place, an Ark usually found in Calculation under this denomination.

To finde these Arks by Calculation.

[Page 45] And this is the first thing Master Gunter and others finde; for South Recliners North Incliners the one being the upper, the other the under face, get the difference between this Ark and the Latitude of the place, the complement of the said residue, remainder, or difference, is the new Latitude sought; but for North Recliners or South In­cliners, the difference between this fourth Arch and the complement of the old Latitude is the new Latitude.

To finde the new Declination:

As the Radius, Is to the Cosine of the Re-Inclination,

So is the sine of the old Declination, To the sine of the new.

This method is hinted to us in Mr. Fosters Posthuma, also in his Book of Dyalling in Anno 1638, where he refers leaning plains to such a Latitude wherein they may become East or West Recliners, but that method is to be deserted, as multiplying more proportions then this, and doth not afford that instrumental ease for pricking down the hours that this doth.

Affections determined.

Such South Recliners, whose meridional Arch is less then the Lati­tude, pass beneath the Pole, and have the North Pole elevated above them, but if the meridional Ark be greater then the Latitude, they pass above the Pole, the North Pole is elevated on the under face, all other affections are before determined. If the meridional Arch be equal to the Latitude, the plain is a Polar plain; for plains leaning Southwards, if the meridional Arch be equal to the complement of the Latitude, the plain is an Equinoctial plain, if it be more, the plain hath less Reclination then an Equinoctial plain, if it be less it hath more, and all affections necessary for placing (and Calculating) the meridian line were before determined.

This method of Calculation findes the Substiles distance from the meridian, not from the plains perpetdicular, wherefore it must be shewed how to place it in Plains leaning Southwards, for plains leaning Northwards use the former directions.

To place the Substile in North Recliners.

In these plains the Meridian and Substilar are to meet at the Cen­ter, and not being drawn through, will make sometimes an Acute, sometimes an obtuse Angle.

When the Plains Meridional Ark is greater then the Colatitude, they make an Obtuse Angle, in this Case, having first placed the Meridian line, above it prick off the complement of the distance of the Substile from Meridian to a Semicircle.

But when the Meridional Ark is less then the Colatitude, prick off the said distance it self above the Meridian line.

In South Incliners.

When the Plains Meridional Ark is greater then the Colatitude, the Substile and Meridian make an Acute Angle, when it is equal to the Colatitude, they make a right Angle, when it is less then the Colatitude they make an Obtuse Angle, and must be prickt off by the complement of their distance to a Semicircle, the Substile al­ways lying on that side of the Meridian, opposite to the Coast of Declination.

A fourth Method of Calculation for leaning Plains. An Advertisement.

In this method of Calculation for all Plains leaning Northward, both upper and under side their Declination is the Arch of the Hori­zon between the North and the Azimuth of the plains South Pole, so that their Declination is always greater then a Quadrant; But for all Plains leaning Southwards, both upper and under face, their Declination is the Arch of the Horizon between the North and the Plains North Pole, wherefore it is always less then a Quadrant; in this sense Declination is used in the following Proportions.

Again,

Get the sum and difference of the fourth and seventh Arch, then if the Colatitude be greater then the complement of the Reclinati­on, the sum is the Substiles distance from the plains perpendicular, and the difference the Inclination of Meridians.

But if it be less, the difference is the Substiles distance from the Plains perpendicular, and the sum the Inclinations of Meridians.

To place the Substile.

For Plains leaning Southwards, when the Angle of the Substile from the Plains perpendicul is less then a Quadrant, it will on the up­perface lye above that end of the Horizontal line that is opposite to the Coast of Declination, and on the under face lye beneath it, but when it is greater, it will lye below the said end, on the upper face, and above it on the under face, but this will not be till the Re­clination be more then the complement of the Reclination of a Polar plain that hath the same Declination; for plains leaning North­wards the Directions of the first Method suffice.

To place the Meridian.

Either Calculate it, and place it according to the directions of the first and second Method, or else Calculate it by this Proportion.

For Plains leaning Northward, the first directions must serve, but for Southern Plains the second, because the distance of the Meridi­an is Calculated from the Substile supposed to be placed, and here the work is converse to that, for in that we supposed the Meridian placed, and not the Substile.

For the Stiles height.

[Page 48] How much the said Ark being doubled wants or exceeds 90 ^d, is the Stiles height.

In South Recliners, if the said Ark being doubled, is less then 90 ^d, its complement is the elevation of the North Pole, and the Plain falls below the Pole.

But if the said Arch exceed 90 ^d the Plain passeth above the Pole, and the excess is the elevation of the North Pole on the under face of a South Recliner, called a North Incliner, and the affections were determined in the first method where the Declination hath its Deno­mination from that Coast of the Meridian to which the Plain look­eth.

These methods of Calculation may not precisely agree one with another, though all true, unless the parts Proportional be exactly Calculated from large Tables in every Operation, which to do as to the Examples in this Book, my leisure would not permit; This last method is derived also from the former Oblique Triangle, the Proportions here applyed, being demonstrated in Trigonometria Brittanica by Mr. Newton.

The Demonstration of the former Proportions

In projecting the Sphere, it is frequently required to draw an Arch through any two different Points within a Circle, that shall divide the said Circle into two equal Semicircles

Construction.

1. Draw a line from one of the given points through the Center, for conveniency through that point which is most remote.

2. From the Center raise a Radius perpendicular to that line.

3. And from the said point draw a line to the end of the Radius.

4. From the end of the Radius raise a line perpendicular to the line last drawn, and where it intersects the former line drawn through the first point and Center, is a third point given, describe a Circle through these three Points, and the Proposition will be ef­fected.

Example.

Let it be required to draw the Arch of a Circle through the two points E and F that shall divide the Circle BD into two equal parts.

[Page 49]

Operation.

From E draw EG, through the Center A, make AD perpen­dicular thereto, joyn ED, and make DG, perpendicular to ED cut­ting EG in G, through E, F, and G, draw the Arch of a Circle which will divide the Circumference BDC into two equal parts in B and C, that is, if CA be drawn, it will pass through B, if not, let it pass above or below, as let it pass below and cut BFE in H.

Demonstration.

By construction EDG and DAG are right Angles; therefore □ AD = ▭ EAG by 13 Prop. 6 Euclid. because EDG being a right Angle, AD is a mean Proportional between EA and AG, but ▭ EAG should be = ▭ CAH by 35. Prop. of 3 Euclid. therefore ▭CAH = □ AD.

But □ AD = □ AI that is = ▭ CAI, therefore ▭ CAH = ▭ CAI which is absurd, therefore CI cannot pass below B, the same absurdity will follow if it be thought to pass above it, there­fore CA produced, will fall in the point B, wherefore BDC is a Semicircle, which was to be proved: And hereof I acknowledge I [Page 50] have seen a Demonstration by the Learned teacher of the Mathe­maticks, Mr. John Leak, to this effect.

To project the Sphere and measure off the Arks of an upright Decliner.

Upon Z as a Center, describe the Arch of a Circle, and cross it with two Diameters at right Angles in the Center, whereto set NESW to represent the North, East, South and West.

Prick off the Latitude from N to L, and lay a Ruler to it from E, and where it cuts NZ, set P to represent the pole.

Prick off the Declination of the plain from E to A, and from S to D, and draw the Diameter AZB, which represents the plain, and DZC, which represents the Poles thereof.

Through the three points CPD, draw the Arch of a Circle, and there will be framed aright Angled Triangle ZHP right Angled at H, in which there will be given the side ZP the complement of the Latitude, with the Angle PZH the complement of the Declination.

Whereby may be found the Stiles height represented by the side PH, the Substiles distance from the plains perpendicular represen­ted by ZH, and the Angle between that Meridian which makes right Angles with the plain, and the Meridian of the place represen­ted by the Angle ZPH, shewing the Arch of Time between the Substile and meridian, called the Inclination of Meridians, from which Triangle are educed those proportions delivered for upright Decliners.

To measure off these Arks. 1. The Substile.

A Ruler laid from D to H, findes the point F in the Limbe, and the Arch CF is the measure of the Substiles distance from the meri­dian, to wit, 21 ^d 41′.

2. The Stiles height.

Set off a Quadrant from F to G, lay a Ruler from G to D and where it intersects BZ, set ☉ which is the Pole of the Circle CPD, lay a Ruler from ☉ to P, and it intersects the Limbe at I, so is the Arch AI the measure of the Stiles height, to wit, 32 ^d 32′

3. Inclination of Meridian.

Lay a Ruler from P to ☉, and it intersects the Limbe at T, and

[Page 51] the Arch WT is the measure of the Inclination of meridians, to wit, 36 ^d 25′.

4. Angle of 12 and 6.

In like manner we may draw a Circle passing through the points WPE, as the prickt Arch PE doth, then in the Triangle ZPQ right Angled at P, we have the side ZP given, and the Angle PZQ to finde the side ZQ, lay a Ruler from D to Q, and you will finde a Point in the Limbe, the distance whereof from C is the measure of the Arch sought, to wit, 57 ^d 49′, to be measured by projection as the Inclination of Meridians.

Lastly the hour-lines, these are represented by meridians drawn through the Poles of the World, as in the Triangle HPQ there will be given the Stiles height PH, and the Angle HPQ, to wit, the Ark of difference between the Inclination of Meridians and the hour from noon, for all hours on that side of the meridian the Sub­stile falls, but on the other side the sum of these two Arks, and this Angle is called the Angle at the Pole; the side required is HQ, the distance between the Substile and the hour line proposed, which must be any hour, though in this Scheam it represents the horary di­stance of six from the Substile.

To project the Sphere for an East or West Reclining or In­clining Plain, Latitude 51 degrees, 32 minutes, a West Plain, Reclining 50 degrees.

Having drawn the Fundamental Scheam, and therein set off the Latitude as before, count the Reclination from N to R, and by lay­ing a Ruler, finde the point A, through it and the North and South points draw the Arch of a Circle which shall represent the plain, finde the pole thereof by setting off a Quadrant from R to G, then through the pole of that Circle ☉, and the pole of the World P, draw the Arch ☉ PR, then in the Triangle NHP right angled at H, we have given the side NP the Latitude, and the Angle PNH the Reclination, to finde PH the Stiles height, and NH the Substiles di­stance from the Meridian, and the Angle NPH the Inclination of Meridians, which is also represented by the Angle BPS.

1. To measure the stiles height.

Set off a Quadrant from B to C, and draw a Line through the Center, and where it intersects the plain at F, is the Pole of the plains Meridian, lay a Ruler from F to H, and it cuts the Limbe at I.

Also lay it thence to P, and it cuts the Limbe at K, the Arch IK is the measure of the Stiles height, to wit, 36 ^d 50′

2. The Substile.

A Ruler laid from ☉ to H, cuts the Limbe at M, and the Arch NM is the measure of the Substiles distance from the Meridian, to wit, 38 ^d 59′.

3. The Inclination of Meridians.

Set off a Quadrant from K to O, and lay a Ruler from it to F, and it intersects the Meridian of the plain at Q, then a Ruler laid from P to Q, findes the point T in the Limbe, and the Arch ST is the measure of the Inclination of Meridians, to wit, 53 ^d 26′.

Otherwise with less trouble lay a Ruler from P to F, and it inter­sects the Limbe at V, and the Arch EV is the measure of the Inclina­tion of Meridians, as before.

To project the Sphere to represent a Declining Reclining Plain. A South Plain Declining 40 degrees East, Reclining 60 degrees, Latitude 51 degrees, 32 minutes.

Having drawn the fundamental Circle, prickt off the Declination, and found the pole point as before, prick the Reclination from B to I, and laying a Ruler to it from A, finde the point R, and through the three points BRA describe a Circle, representing the plain, also from K finde the pole thereof ☉, and through the two points P and ☉ draw the Arch of a Circle FG, representing the plains Meridian, at the intersection of the plain with the Meridian set Z, and draw the Arch of a polar plain through P to S, and there will be several Triangles Constituted, from which were derived the several Methods of Calculation.

In the right Angled Triangle ANZ there is given NA the comple­ment of the Declination, and the Angle NAZ the complement of the Reclination, whereby may be found ZN the plains meridional Ark, which taken from NP rests ZP the complement of the new Lati­tude; [Page 53] also the Angle NZA which is the complement of the new De­clination, and hence were derived the Proportions for the third Method, likewise in the same Triangle may be found ZA the meri­dians distance from the Horizon.

In the Oblique Angled Triangle CP☉, there is given the side CP, the complement of the Latitude, the side C☉, the complement of the Reclination with the Angle PC☉, the complement of the Decli­nation from the South to a Semicircle, whereby may be found the Angle CP☉, the Inclination of Meridians, and the Angle C☉P whereof the measure is RH the distance, of the Sub­stile from the Plains perpendicular, and the third side P☉, the com­plement whereof is PH the Stiles height, and from hence was derived the third method of Calculation suited to Proportions for finding both the unknown Angles of an Oblique Spherical Triangle at two Operations, when there is given two sides with the Angle compre­hended between them.

The first method of Calculation is built upon the perpendicular Trigonometrie, for the perpendicular PS reduceth the former Ob­lique Triangle, to two right Angled Triangles, to wit, the right An­gled Triangle, PSC, and the right Angled Triangle PS☉, both right Angled at S.

In the right Angled Triangle PSC, we have CP given the Cola­titude, and the Angle PCS the Declination to finde SC a Polar plains Reclination thereto.

Again,

In Oblique Spherical Triangles, reduced to two right Angled Triangles by the demission of a perpendicular, it is a common in­ference in every book of Trigonometry, when two sides with the Angle comprehended are given, to finde one of the other Angles: That,

And so in that Oblique Triangle, the difference between the Reclination of the plain proposed, and the Polar plain is RS, then because R☉ is a Quadrant, S☉ is the complement of the former [Page 54] Ark, therefore it holds: As s SO: t SCP ∷ s SC: t S☉P which is the very proportion delivered delivered in the said Method for finding the Substiles distance.

Then in the right Angled Triangled PS☉, we have S☉, and the Angle S☉P to finde the side P☉, whereby is got the Stiles height; the Inclination of Meridians is found in the Oblique Spherical Tri­angle by the proportion of Opposite sides and Angles.

Lastly, in the right Angled Triangle ZRC, there is given RC, and RCZ to finde RZ the distance of the Meridian from the plains per­pendicular.

To measure the respective Arks abovesaid.

The Polar Reclination CS is 31 ^d 19′.

The Scheam determineth all the affections of the plain.

1. It shews that H the point of the Substilar lies on that side the plains perpendicular, that is towards the Coast of Declination.

2. That Z the point for the place of the Meridian, lyes towards the same Coast as before, but below the Substilar Line.

3. The Arch PH shews you that the North Pole is elevated above the upper or Reclining face.

After the same manner may all the Requisite Arks be measured, and affections determined for all plains whatsoever.

A South Plain Declining 30 degrees Eastwards, Reclin­ing 25 degrees, or a North Plain Declining 30 de­grees Westwards, Inclining [...] degrees.

In this Scheam we have the same Oblique Triangle PC☉ reduced to two right Angled Triangles PSC and PS☉, SC is the Inclination of a Polar plain, and RC the Inclination of the plain proposed, the dif­ference is SR, and the complement of it, is the complement of S☉ to a Semicircle, because S☉ is greater then a Quadrant, and the pro­portions are wholly the same, though the Triangle have sides great­er then a Quadrant.

The North Pole is elevated on the Inclining face, the Meridian Z lyes from the plains perpendicular towards that end of the Horizon­tal Line, opposite to the Coast of Declination, the same way and be­neath it lyeth the Substilar.

The complement of the new Latitude ZP is 10 ^d 10′

The new Declination, viz. the complement of NZA is 26 ^d 57′

The meridians distance from the Plains perpendicular RZ 13 ^d 43′

The Substiles distance therefrom RH 18 ^d 22′

The Stiles height PH is—9 ^d—3′

The Inclination of Meridians, to wit, the Angle CP☉ 27 ^d 18′

The Polar Reclination—CS—34 ^d 31′

A South Plain Declining 40 degrees East, Inclining 15 degrees, or rather a North Plain Declining 40 de­grees West, Reclining 15 degrees.

Here again the Oblique Triangle CP☉ is reduced to two right Angled Triangles PSC and PS☉, and SR is the sum of the Polar Reclination SC, and the Re Inclination of the Plain proposed CR, and S☉ is the complement hereof, because ☉R is a Quadrant, finde the Equinoctial point AE.

The Polar Reclination CS 31 ^d 19′.

The new Latitude ZAE is 32 ^d 15′.

The new Declination being the complement of NZA is 38 ^d 23′

The Meridians distance from the plains perpendicular ZR 12 ^d 33′

The Substiles distance from the plains perpendicular RH 32 ^d 16′

The Stiles height PH is—41 ^d 30′

The Inclination of Meridians ☉PC rather the Acute Angle ☉PN is 55 ^d 58′

The Meridian Z lies from the plains perpendicular towards the Coast of Declination on the Reclining side, but must be drawn through the Center, because the Sun at noon casts his shadow North­wards, unless in the Torrid or Frozen Zone, and the Substile Hlyes on the other side the plains perpendicular.

A North Plain Declining 40 degrees Eastwards, Reclining 75 degrees.

In this plain likewise the Oblique Triangle C☉P is reduced to two right Angled Triangles PS☉ and PSC by the perpendicular PS which is part of the Arch of a polar plain, here CR more CS is equal to the sum of the Plains Reclination proposed, and of the Polar plains Reclination, which is greater then a Quadrant for the Arch R☉ is a Quadrant; now the Cosine of an Ark greater then a Qua­drant is the Sine of that Arks excess above a Quadrant, wherefore the Sine of S☉ is the Cosine of the sum of both the Reclinations, and the Case the same as before.

The Arch RH is the measure of the Substiles distance from the [Page]

[Page] [Page 59] Plains perpendicular, which being greater then a Quadrant, shews that the Substile must lye below the end of the Horizontal Line opposite to the Coast of Declination, to wit, it must be depressed 32 ^d 39′ measured by the Arch KT, the complement of TA its distance from the plains perpendicular, the point T being found by laying a Ruler from ☉ to H, which may also be measured from F, and will fall above the Horizontal Line D, but then must be drawn through the Center.

Prick off a Quadrant from G to X, and draw XC, it cuts the plain at Y, a Ruler laid from Y to H, and P findes the points 2, 3 in the Limbe, and the Arch 2, 3 being 61 ^d 31′, is the Stiles height; orther complement thereof to a Semicircle might be found by measuring the Arch PF.

A Ruler laid from ☉ to Z, findes the point V in the Limbe for the Meridian Line, from which draw a Line through the Center on the other side, and it will be placed in its true Coast and quantity from the Plains perpendicular at A, to wit, 39 ^d 2′.

The Inclination of Meridians, to wit, the Angle CP☉ is 20 ^d 30′,

The new Latitude ZAE 26 ^d 38′.

The new Declination is 9 ^d 35′ to wit, the complement of KRB.

The Polar Reclination CS is 31 ^d 19′.

The truth of this Stereographick Projection is fully handled by Aguilonius in his Opticks, and how to determine the affection of any Angle of an Oblique Spherical Triangle, I have fully shewed in a Treatise, called the Sector on a Quadrant.

For the Resolution of Spherical propositions, Delineations from proportions or the Analemma, will be more speedy and certain (though they may also be thus resolved) which I have handled at large in the Mariners Plain Scale new plain'd.

To determine what hours are proper to all kinde of Plains.

To do this it will be necessary to project upon the Plain of the Horizon, the Summer and Winter Tropicks.

Get the Sum and difference of the Colatitude, and of the Suns greatest Declination, so we shall obtain his greatest and least Meri­dian Altitudes.

The depression of the Tropick of Cancer under the Horizon, is [Page 60] equal to the least Meridian Altitude, and the depression of the Tro­pick of Capricorn to the greatest. Example:

• 38 ^d 28′ Colatitude
• 23 ^d 31′
• 61 ^d 59′ greatest 14 ^d 57′ least
  • Meridian Altitude, having drawn the Primitive Circle, &c. as before.

Prick 14 ^d 15′ from S to C, and 61 ^d 59′ from S towards W, a Ru­ler laid from the points found, will intersect the meridian ZS at the point L for the Winter Tropick, and K for the Summer Tropick, through which the Circles that represent them are to pass, to finde the Semidiameters whereof, set off their depression from N to­wards E, thus 14 ^d 57′ the depression of the Summer Tropick ter­minates at O, a Ruler laid from E to ☉, findes the point X in the meridian SZN produced, so is XK the Diameter of the Summer Tropick, which being divided into halfs, will finde the Center thereof whereon to describe it.

In like manner is the Diameter of the Winter Tropick to be found; or if the Amplitude be given (or found as elsewhere is shew­ed) which at London is 39 ^d 54′, and set off both ways from G and E we shall have three points given through which to draw each Tropick, and the Centers falling in the Meridian Line will be found with half the trouble, as to finde a Center to three Points.

Also Project the Pole Point P as before, being thus prepared FKG will represent the Summer and HLI the Winter Tropick.

Let it be required to know what hours are proper for a South plain Declining 30 ^d Eastwards, through the three points BPA describe the the Arch of a Circle BQP, then laying a Ruler from B to Q, finde the point R in the Limbe, and from it set off a Quadrant to M, then a Ruler laid from B to M findes the point ☉ the Pole of the hour Circle BQP, then laying a Ruler from P to ☉, it findes the point T in the Limbe, and the Arch ET being 65 ^d 40′ is the measure of the Angle BPS, which turned into Time is 4 ho. 23′ prope, and sheweth that at no time of the year the Sun will shine longer on the South side of this plain, then 23 minutes past 4 in the afternoon.

In like manner if the Arch of a Circle be drawn through the two points PV, we may finde the time when the Sun will soonest in the morning begin to shine on the South side of this plain.

The Angle QPZ may be found by Calculation in the right Ang­gled [Page] [Page]

[Page 61] triangle PQZ, in which besides the right Angle at Q, the side PZ and the Angle QZP, are given but to measure it off by Projection will be sufficient for the Dyallists use.

So if there were a South plain Declining 60 ^d Eastwards, Reclining 40 degrees here represented by BRA, if it were re­quired to know what hours are proper for the upper, and what for the under face, then where the plain intersects the Tropicks as at I and K, draw two Meridians into the Pole at P, to wit, IP and KP, and first finde the Angle IPZ, as was before shew­ed, to wit, 53 ^d 14′ which in time is 3 hours 33 minuutes, shewing that the Sun never shines longer on the upper face of the Plain, then 33 minutes past 3 in the afternoon, which is capable of receiving all hours from Sun rising to that period of time, and the Angle KPZ, to wit, 39 ^d 50′ in time 2 hours 39 minutes, shews that the Sun never begins to shine sooner on the under face then 39 minutes past 2 in the afternoon after, which all the hours to Sun-set may be expressed.

To finde these Arks of time by Calculation, there must be given the Stiles height above the Plain 15 ^d 22′, PH, and the complement of the Inclination of Meridians to a Semicircle, 136 ^d 32′, to wit, HPS, then in the right Angled Triangle PHI there is given PH the Stiles height, PI the complement of the Declination, besides the right Angle at H, to finde the Angle IPH 83 ^d 18′ which taken from the com­plement of the Inclination of Meridians HPS, there rests the An­gle IPS the Arch of time sought, to wit, 53 ^d 14′.

The ascensional difference may be found by drawing the Arch of a Circle through the three points TPF, and thereby the length of the longest day determined that no hours be expressed, on which the Sun can never shine.

Another manner of Inscribing the hour-lines in all Plains having Centers.

The method here intended, is to do it in a parallelogram from the Meridian line, whence the hour-lines may be prickt down by a Tangent of three hours, with their halfs and quarters from a Sector, without collecting Angles at the Pole, or by help of a Scheam which I call the Tangent Scheam, the foundation of this Dyalling suppo­seth the Axis of the world to be inscribed in a Parallelipiped on con­tinued about the Axis, the sides whereof are by the plains of the re­spective [Page 62] hour Circles in the Sphere divided into Tangent-lines, that is to say, each side is divided into a double tangent of 45 ^d set toge­ther in the middle, and the said parallelipiped on being cut by any Plain, the end thereof supposed to be intersected, shalbe either a right or Oblique Angled parallelogram, and then if from the opposite tan­gent hour points on the sides of the intersected parallelipipedon, lines be drawn on the Plain, they shall cross one another in a Center, and be the hour-lines proper to the said plain, but of the Demonstrati­on hereof, I shall say no more at present, the inquisitive Reader will finde it in the Works of Clavius.

To draw the Tangent Scheam.

I have before in Page 10 shewed how to divide a Tangent Line into hours and quarters, which in part must be here repeated; draw any right Line, as MABH, from any point therein as at B, raise a per­pendicular, and upon B as a Center, describe the Quadrant CH, and prick the Radius from B to A, from C to G, from H to F, and lay­ing a Ruler from A to F, and G, you will finde the points D and E upon the perpendicular CB, I say the said perpendicular is divi­ded into a Tangent line of three hours, and the halfs and quarters may be also divided thereon, by dividing the Arches CF, FG, and GH into halfs and quarters, then from those subdivisions, laying a Ruler to A, the halfs and quarters may be divided on, as were the whole hours.

Being thus prepared, draw the Lines MB, LE, KD, and IC, all parallel one to another, passing through the points B, 1, 2, 3.

[Page 63] In this Scheam they are perpen­dicular to BC, but that is not material, provided they pass through the same points, and are parallel one to another, yet notwithstanding the points A and H must be in a right Line perpendicular to CB. This Scheam thus prepared, I call the Tangent Scheam, because a Line ruled any way over it, shall be divided also into a Tangent of the like hours and quarters, whence it follows that one of these Scheams may serve to inscribe the hour-lines into many Dyals, which I shall next handle.

To inscribe the Hour lines in a Horizontal Dyal.

Having drawn the Meridian line M, XII, and perpendicular thereto the hour-line of six, Let it be observed that the sides of the Horizontal Dyal in page 8. to wit VI, IX, and VI, III and IX, III, are a right Angled Parallelogram, the one side whereof being the Diameter of the Circle being Radius, the other side thereof must be made equal to the Sine of the Latitude, in that Scheam the nearest distance from L to MF was the Sine of the Lati­tude or Stiles height, the Semidiameter of the inward Circle, being Radius, and that extent being doubled and pricked from M to VI on each side, as also from F twice to IX and III, by those extents the Pa­rallelogram was bounded, the side III, F IX, being parallel to the Horizontal Line.

Then take the extent MF from the Horizontal Dyal, and place it [Page 64] in the Tangent Scheam from M to O, and draw the Line MO, and the respective Divisions of the said Line being cut by the parallels of the Tangent Scheam are the same with the Divisions of the hour-lines on the inward sides of the Horizontal Dyal VI, IX, and VI, III, and from the tangent Scheam they are to be transferred thither with Compasses.

Also place F, IX or F, III from the Horizontal Dyal into the Tangent Scheam from M to N, and draw the line MN, which being cut by the parallels of the tangent Scheam, the Distances of those Divisions from M are to be pricked down in the Horizontal Dyal, the first from F to XI, and I the second from F to X and II, &c.

To inscribe the hour-lines in a direct erect South Dyal.

The Diameter of the Circle in the South Dyal in page 8, is the same as in the Horizontal, and the nearest distance from L to FM was the Consine of the Latitude, and was pricked twice on the Ho­rizontal Line from M to VI on each side, whereby that inward Pa­rallelogram was limited.

Wherefore the divisions of the Line MO in the Tangent Scheam, are the same with the hour distances in this Dyal on the sides VI, III, and VI, IX; then for the divisions of the hours on each side of XII, take the extent XII, IX or XII, III, and because it is less then the outward parallel distance of the sides of the Tangent Scheam, ha­ving therein made MI perpendicular to BM, place this extent from I to P, and draw the line MP, then prick the extent RL from the Tangent Scheam, from F to XI and I, and the extent QK, from F in the Dyal, to X, and II, and from the hour points so found, draw lines into the Center at M, and they shall be the hour-lines required.

To delineate an upright Decliner in an Oblique Parallelogram.
An upright South Dyal Declining 30 degrees West, Latitude 51 degrees, 32 minutes.

First draw the Meridian or Plains perpendicular CN, and upon C as a Center, with the Radius of the Dyalling Scheam, describe

[Page]

[Page 65] an Occult Ark, wherein prick off NT from the aforesaid Scheam, and draw TC for the Substile; also set off in the said Ark TK equal to TK of the beforementioned Scheam, and draw KC for the Stile; perpendicular to the Meridian passing through the Center C, draw the Plains Horizontal Line, and therein from the Center prick down from C to F, the Cotangent of the Latitude, to wit, BF upon which the Sine of the Declination is to be erected to the same Radius, and the nearest distance from A to DV in that Scheam, is the Sine of the declination, which being prickt on the perpendicular FG, draw a Line from the point G into the Center, and it shall be the hour-line of six.

In Latitudes under 45 ^d the side of the square AV must be assumed to be the Cotangent of the Latitude, the Radius whereto will be AF the Tangent of the Latitude, to which Radius being prickt on the side of the square from the Center, the Sine of the declination must be taken out as before, and erected on the Cotangent of the Lati­tude, and this work must be performed on that side of the Center on which the Substile lyes.

To fit in the Parallelogram.

Produce the Lines WE, VD, and VL, in the Dyalling Scheam far enough, then assuming any extent to be Radius, enter it on the lines VD and VL from the Center to Y and Z, the nearest distance from Y to VE is the Cofine of the Latitude to that Radius which enter on the Line of 6 or GC, so that one foot resting thereon, the other turned about may just touch CN, at the point found set H, and make CI on the other side equal to CH.

The nearest distance in the Dyalling Scheam from Z to VE, is the Cosine of the declination to the former Radius, which prick on the Meridian line from C to L.

And draw a Line through L parallel to HI, and therein make LP, LQ each equal to CH, and draw HP and IQ, and there will be an oblique Parallelogram constituted, the sides whereof will be Tan­gent Lines.

Nota, we might assume the point G in the hour-line of six, for the Parallelogram to pass through, and the nearest distance from D to VE in the dyalling Scheam, would be the Cosine of the declinati­on [Page 66] to the same Radius to be prickt on the Meridian Line as be­fore.

To inscribe the hour-lines.

In the following Tangent Scheam made as the former, produce CB, and make BQ equal to BC, then take the extent PQ on the dyal, and upon Q as a Center describe the Ark Y therewith, and draw the Line YC just touching the extreamity, then you may pro­portion out the hours in this manner.

The nearest distance from D to YC in the Tangent Scheam, being prickt on the side of the Parallelogram from P and Q, will finde the points 10 and 2, through which those hour-lines are to pass, also the nearest distance from E to YC, being there prickt, findes the points 11 and 1 as before, otherwise take the extent PL from the Dyal, and in the Tangent Scheam upon C as a Center, describe the Ark N therewith, and draw BN just touching the outward extreamity of that Ark, the nearest distance from E to BN, prick in the Dyal from L to 11 and 1, and the nearest distance from D to BN, place from L to 10 and 2, thus you have found points through which part of the hour-lines are to be drawn into the Center at C; or having found the points P and Q on the plain, if it be desired to inscribe the hours without drawing the side of the Parallelogram PQ, which must also be performed without the help of Compasses, though they be at hand, then in the Tangent Scheam prick the extents E and D, from B to I and K, and through the respective points C, D, E, B, I, K, Q, draw lines all parallel one to another, making any Angle at pleasure with the said Line, I say then that any line ruled over those parallels, shall be divided into a line of double tangents, as is LM, which is made [Page 67]

[Page 68] equal to PQ, the side of the Parallelogram in the former Dyal, if therefore you double the Paper in the line LM, and apply the points L, M to the points QP on the plain, you may by the edge of this Paper Tangent Scheam so doubled, prick down the hour points on the plain, and lines drawn through them into the Center, shall be the hour-lines required, and making BO equal to the sides of the Parallelogram HP or IQ, and it shall contain the hour distances to be transferred into the said sides.

If a line be drawn in the Dyal from H to L, and from L to I, the hour-lines being drawn shall divide each of these lines into a double Tangent, and consequently the hour-lines may also be prickt off on the said lines, after the method now prescribed.

For upright far Decliners and such plains as have small height of Stile, recourse must be had to former directions for drawing them with a double Contingent line, each at right Angles to the Substile. The foundation whereof is this, Any point being assumed in the Sub­stilar line of a Dyal, the nearest distance from that Point to the Stile, is the Sine of the Stiles height, the Radius to which Sine is the di­stance of the assumed point in the Substilar line from the Center of the Dyal; Then in all Dyals the hour distances from the Substilar line are Tangents of the Angle of the Pole, the Sine of the Stiles height being made the Radius thereto, having finished the delineati­on of the Dyal, the Stile is to be placed directly over the Substilar line, without inclining to either side of the Plain, making an Angle therewith equal to its height above the same, for the Substilar line is elsewhere defined to be such a line over which the Stile is to be pla­ced in its nearest distance from the Plain, therefore if the Stile in­cline on either side, it will be nearer to some other part of the Plain then the Substilar line, whence it comes to pass in places near the Equinoctial, if an upright Plain decline but very little, the Substile is immediately cast very remote from the Meridian.

Another way to prick down the hour-lines in Declining leaning Plains.

Every such Plain in some Latitude or other will become an upright Decliner: First therefore by the former directions prick off the Sub­stile, Stile and Meridian, in their true Coast and quantity, and perpen­dicular to the Meridian, draw a line passing through the Center, and [Page]

[Page] [Page] [Page]

[Page 89] it shall represent the Horizontal line of the Plain in that new Lati­tude as here VS; from any point in the Stile as K, let fall a perpen­dicular to the Substile at I, and from the point I, in the Substile let fall a perpendicular to the Meridian at P.

To finde the new Declination.

Prick IP on the substilar line from I to R, and draw RK, so shall the Angle IRK be the complement of the new declination, and the Angle IKR the new Declination it self.

To finde the new Latitude.

Upon the Center V with the Radius VK, describe a Circle, I say then that VP is the Sine of the new Latitude to that Radius which may be measured in the Limbe of the said Circle, by a line drawn pa­rallel to VS, which will intersect the Circle at F, so is the Arch SF the measure of the new Latitude.

To prick off the Hour-line of Six.

This must be prickt off below the Horizontal line, the same way that the substilar lyes: The proportion, is,

If RK be Radius, then is VP the Tangent of the new Latitude, but if we make VP Radius, then is RK the Cotangent of the new Latitude.

Wherefore prick the extent RK on the Horizontal line from V to N, and thereon erect the Sine of the Declination to the same Radi­us perpendicularly, as is NA, and a line drawn into the Center shall be the hour-line of six; the proportioning out of the Sine of the Declination to the same Radius, will be easily done, enter the Radius VP from K to D, and the nearest distance from D to IK, shall be the Sine of the new declination to that Radius.

To fit in the Parallelogram.

This is to be done as in upright Decliners, for having drawn a line [Page 70] from the new Latitude at F into the Center, if any Radius be en­tred on the said line from V, the Center towards the Limbe, the nearest distance from that point to VP the Meridian line, shall be the Cosine of the new Latitude to that Radius.

Again if the same Radius be entred on RK produced if need be, the nearest distance to VR (produced when need requires) shall be the Cosine of the new Declination, and then the hour-lines are to be drawn as for upright decliners, nothing will be doubted concern­ing the truth of what is here delivered, if the demonstration for in­scribing the Requisites in upright decliners be well understood, it being granted that Oblique Plains in some Latitude or other will be­come upright decliners.

There are two Examples for the Latitude of London suited to these directions, in both which the Letters are alike, the one for a South Plain declining 40 ^d Eastwards, Inclining 15 ^d, the other for a North Plain declining 40 ^d East, Reclining 75 ^d.

To finde a true Meridian Line.

For the true placing of an Horizontal Dyal, as also for other good uses it will be requisite to draw a true Meridian Line, which proposition may be performed several ways, amongst others the Learned Mathematician Francis van Schooten in his late Miscellanies demonstrates one, performed by help of three shadows of an up­right Stile on a Horizontal Plain, published first without Demonstra­tion in an Italian book of dyalling by Mutio Oddi.

Let AB, AC, and AD represent 3 shadows made on a Horizontal plain in one day, by the shadow of the stile or wyre AE, erect­ed perpendicularly from the point A, of which shadows if two be found e­qual, a line let fall perpendicularly [Page 71] from the Point A, upon a line joyning, the extremities of those two shadows, shall be a true Meridian line.

But if all three be unequal, as let AC be the least, erect three lines from the point A, perpendicular to AP, AC, AD as is AF, AG, and AH equal to the Stiles height AE, and draw lines from the ex­treamities of the three shadows to these three points as are FB, GC, and HD; then because AC is less then AB, therefore GC will be less then FB, by the like reason GC will be less then HD, wherefore from FB and HD cut off or Substract FI and HK equal to GC, and from the points I and K let fall the perpendiculars IL, KM, upon the Bases AB, AD, afterwards draw a line joyning the two points M, L, and from the said points let fall the perpendiculars LN equal to LI, and MO equal to MK.

Then because the two shadows AB and AD are unequal, in like manner FB and HD will be unequal; but forasmuch as FI and HK, are equal by construction, it follows that LI, KM, or LN and MO will be unequal, and forasmuch as these latter lines are parallel a right line that connects the points O and N, being produced will meet with the right line that joyns M, L produced, as let them meet in the point P, from whence draw a Line to C, and it shall be a true line of East and West, and any Line perpendicular there­to shall be a Meridian line, thus the perpendicular AQ let fall there­on, is a true Meridian Line passing through the point A, the place of the stile or wyre.

Whereto I adde that if MO & LN retaining their due quantities be made parallel it matters not whether they are Perpendicular to ML or no, also for the more exact finding the Point P, the lines MO and LN, or any other line drawn parallel to them, may be multipyled or increased both of them the like number of times from the points M and L upwards, as also from the points O and N downwards, and Lines drawn through the points thus discovered, shal meet at P with­out producing either ML or ON. See 15 Prop. of 5 Euclid. and the fourth of the sixth Book.

The greater part of van Schootens Demonstration is spent in pro­ving that ML and ON produced will meet somewhere, this for the reasons delivered in the construction I shall assume as granted, then understand that the three Triangles ABF, ACG and ADH stand perpendicularly erect on the plain of the Horizon beneath them, up­on [Page 72] the right Lines AB, AC and AD, whence it will come to pass that the three points F, G, and H meet in one point, as in E the top of the Stile AE, and that the right lines FB, GC and HD are in the Conique Surface of the shadow which the Sun describes the same day by his motion, the top of which Cone being the point E.

Wherefore if from those right Lines we substract or cut off the right Lines FI, GC and HK being each of them equal to one ano­ther, then will the points I, C and K, fall in the circumference of a Circle, the plain whereof is parallel to the plain of the Equator; and therefore if through the points K and I such a Position a right Line be imagined to pass, and be produced to the plain of the Hori­zon, it will meet with ML produced in the point P, where ON be­ing produced, will also meet with it; so that the point P being in the Plain of the Circle, as also in the plain beneath it, as also the point C being in each Plain, a right line drawn through the Points P and C will be the Common Intersection of each plain, and the line PC will be parallel to the Plain of the Equator, and is therefore a true Line of East and West, which was to be proved.

On all Plains though they Decline, and Recline, or Incline, after the same manner may be found the Line PC, which will represent the Contingent Line of any Dyal, and a perpendicular raised upon the Line CP shall be the Substilar Line, which in Oblique Plains is the Me­ridian of the Plain, but not of the place, unless they are both Coinci­dent from this manner of finding a Meridian Line on a Horizontal plain, nothing else can be deduced without more Scheams: From the three shadows, may be had the three Altitudes, and the Meridian line being given, the Azimuths to those three shadows are likewise given, which is more then need be required in order to the finding of the Latitude of the place, and the Declination and Amplitude of the Sun, which because this Scheam doth not perform of it self, I shall adde another to that purpose.

By three Altitudes of the Sun, and three shadows of an Index on an Horizontal Plain, to finde a true Meridian Line, and consequently the Azimuths of those shadows, the Latitude of the place, the Suns Amplitude and Declination.

[Page 73]

It will be convenient to take two shadows, when the Sun is on one side of the Meridian, and the third when on the other side▪ and the performance will be most exact when the Sun is near the Summer Solstice

• Let the three shadows be CA, the Altitude whereto is AF 22 ^d, 28′
• The second shadow CB, the Altitude whereto is BG 59 ^d, 21′ both these in the morning,
• The third shadow in the afternoon CD the —
• Altitude whereto is —
  • DH 18 ^d, 20′

Let the Angle ACB be 70 ^d, and the Angle BCD 135 ^d, by follow­ing Operations we shall finde that the shadow CA, is 10 ^d to South­wards of the West, that the shadow CB is 60 ^d to Northwards of the West, that the shadow CD is 15 ^d Southwards of the East.

[Page 74] Having from the three shadows prickt off the three Altitudes to F, G, and H, from those points to the shadows belonging to them, let fall the perpendiculars FI, GK, HL, which shall be the Sines of those Altitudes, and the Bases IC, KC, and LC shall be the Cosines, from the point K in the greater Altitude, draw Lines to the points I and L in the lesser Altitudes and produce those lines.

From the points K and I, the Sines of the two Altitudes, are to be erected perpendicularly, thus KM is made equal to KG, and IN is made equal to IF, then producing MN and KI, where they meet as at W, is one point, where the plain of the Suns parallel of Declina­tion intersects the plain of the Horizon; in like manner on the Base KL, the Sines of two Altitudes KG, and LH, ought to stand perpen­dicularly from the points K and L in their common Base, but if they retain the same height, and are made parallel to one another, a line joyning the points of the tops of those Sines produced, shall meet with the line joyning the points of their Bases produced, in the same point as if they were perpendicular.

Thus KP and LQ are drawn at pleasure through the points K, and L parallel one to another, and KG is the Sine of the greater Altitude, and LO is equal to the Sine of the lesser Altitude, and these two points being joyned with the Line OG produced, meets with the Line KL produced, in the point E, another point where the plain of the Suns parallel intersects the plain of the Horizon.

Or you may double KG, and finde, the point P, as also double LO and finde the point Q, a line drawn through P and Q, findes the point E, as before, but with more certainty.

If you joyn EW it shall be a true Line of East and West, and a perpendicular let fall thereon from C the Center, shall be a true Me­ridian line to the perpendicular Stile or wyer, as is CS.

The Arch SR is the Suns Amplitude from the North 50 ^d 6′ the shadow being contrary to the Sun, casts his parallel towards the South.

Draw KT parallel to SC, and it shall be the perpendicular dist­ance between the Sine of the Suns greatest Altitude, and the Inter­section of his parallel with the Horizon.

Upon K erect KV equal to the Sine of the Suns greatest Altitude of the three KG, and the Angle KTV shall be equal to the comple­ment of the Latitude, for the Angle between the plain of any pa­rallel [Page 75] of declination and the Plain of the Horizon, is always equal to the complement of the Latitude, and if upon T as a Center with the Radius CS, you describe the Ark KX, the said Ark shall measure the complement of the Latitude in this example 38 ^d 28′ which be­ing given together with the Sine of the Amplitude CY, it will be easie to draw a Scheme of the Analemma, whereby to finde the Suns declination, the time of rising, and setting, his height at six the Azimuth thereto, the Vertical Altitude and hour thereto, &c. and many other propositions depending on the Suns motion, as I have elsewhere shewed.

The whole ground hereof is, that a right line extended through the tops of the Sines of any two Altitudes of the Sun taken the same day before his declination very, shall meet with the Plain of the Horizon in such a point where the Plain of the Suns parallel inter­sects the Plain of the Horizon, and finding of two such points, a line drawn through them, must needs represent the intersection of those two Plain; in the former Scheme the Sines of the two Alti­tudes are KM and IN, a line drawn through the bottomes of those Sines, as KI extended shall be in the Horizontal Plain, and the Line MN extended through the tops of those Sines is in the plain of the Suns parallel, as also in the Horizontal plain; now whether these Lines stand erect or no is not material, provided they retain their Parallelisme and due length, and pass through the points of their Bases I, K, for the proportion of the perpendiculars to their Bases, will be the same notwithstanding they incline to the Horizon.

To Calculate the Latitude, &c. from three shadows.

A usual and one of the most troublesome propositions in Spherical Trigonometry, is from three shadows to finde the Latitude of the place: thus Maetius propounds it, and with many operations both in plain and Spherical Triangles resolves it.

The first operations are to finde the Suns 3 Altitudes to those sha­dows, and that will be performed by this proportion,

Next Maetius gives the distances between the points of the three shadows, and then by having three Sides of a Plain Triangle, he findes an Angle, to wit, the differences of Azimuth between the re­spective shadows, which Angle may be measured off the Plain with Chords.

But propounding it thus, Three Altitudes of the Sun above a Horizontal Plain, with the differences of Azimuth between the three shadows belonging to those Altitudes, being given, let it be required to finde the Suns true Azimuth, the Latitude of the place, and the Suns Amplitude.

And how this may be Calculated from the former Scheme, I shall now shew.

In the Triangle ICK of the former Scheme, the two sides IC and CK, represent two shadows, and the Angle ICK is the Angle or difference of Azimuth between them, and the said sides IC and CK, are the Cosines of the Altitude proper to those shadows; now by seven Operations in right lined Triangles, we may finde the proper Azimuth or true Coast of any of those shadows.

1. In the right lined Triangle ICK, having the two sides IC and CK, with the Angle between them ICK, at one operation may be found both the other Angles CIK and IKC.

2. In the same Triangle by another operation, may be found the Side IK.

Then to proceed, draw Nf parallel to IK, and fM will be the difference of the Sines of both the Altitudes belonging to those shadows, whereby may be found KW.

3. The proportion lyes,

• 4.
• 5.
• 6.
  • By three like Operations may be found in the other shadow
  • Triangle, the Angles CKL, and KLC, with the Side KE.

Having proceeded thus far to the Angle IKC, adde the Angle CKL, the sum is equal to the Angle WKE.

7. Then in the Triangle WKE, we have the two Sides thereof WK, [Page 77] and KE given, and the Angle comprehended by them, and at one Operation we may finde both the other Angles EWK, and KEW, the complement of the Angle TWK, is the Angle WKT. The dif­ference between the Angles WKT, and IKC, is the Angle CKZ, which shews the Suns Azimuth from the Meridian proper to that shadow, which may be otherways found, for the difference between EKT and TKC, also shews it.

By two other Operations the Latitude may be found.

By another Operation may be found the Amplitude CY, having found the Angle ZKC, the Angle CKp is the complement thereof, then it holds:

To make any Dyal from three Shadows.

The Geometrical performance of the former Proposition, is insisted upon by Clavius in his Book of the Astrolable, but▪ he mentioneth no method of calculation as derivable from it: from this proposition Monsieur Vaulezard a French Mathematician educeth a general me­thod for making of Dyals, from three shadows of a Gnomon stuck into a wall at randome, whereof he doth not so much as mention any demonstration; I shall endeavour to deliver the method thereof with as much perspicuity as I can.

Every Plain in some place or other, is an horizontal Plain; admit an Oblique Plain in our Latitude, the Substilar line represents the me­ridian of that place, and any contingent line drawn at right Angles, thereto will represent a true line of East and West in reference to that horizon, and if the hours did commence at 12, from each side the substile, the Dyal here would shew the true time of the day there.

In every oblique Plain assuming any Point in the Stile, and cros­sing the Stile with a perpendicular to that Point, which shall meet with the Substile, the Point so found in the Substile, is the Equi­noctial Point, in respect of the assumed Point in the Stile; hence we may inferre that if these two Points and the Substilar line were gi­ven [Page 78] the Center of the Dial might be easily found, now the former construction applyed to an oblique Plain, assumed to be an Horizon­tal Plain, in respect of some unknown place, will find the Equinoctial Point and the substilar Line, the Stile Point being assumed in the extre­mity of a Gnomon any wayes placed or stuck in a wall at randome.

From the former Scheme it may be observed that the Sine of any of the three Altitudes being erected on the Perpendicular between the foot of that Sine and the Suns parallel▪ gave an Angle equal to the elevation of the Equinoctial above the Horizon, which is the thing sought in the following work, but the Sine of the greatest Al­titude performs the proposition best.

If a stick or pin be stuck into a wall for this purpose, and doth not make right Angles therewith, a Perpendicular must be let fall from the extremity thereof into the wall, which is called the perpendicu­lar Stile, and the distances of the shadowes; from the foot thereof must be measured thence, in respect of the tip of this perpendicular Stile the Equinoctial Point must be found, wherefore it will be con­venient to assume the said perpendicular Stile to be the Sine of the greatest of the three Altitudes of the Sun above the Plain, which properly in respect of Us are Angles between the wall and the Sun. Up on this assumption it will follow, that the shortest shadow will be the Cosine of that Angle, and the distance between the tip of the Stile, and the extremity of that shadow will be the Radius: now from the lengths of the three shadows, and the height of the per­pendicular Stile, it will be easie to find the Sines and the Cosines of the Angles between the wall and the Sun.

In the following figure, let the Circle be supposed to be described upon a leaning Plain, on which let AB represent a Stile stuck in at randome, the extremity of the three shadows whereto are the points D, E, F, first find the perpendicular Stile thereto CB. and from C draw lines to the three shadowes: now to find the Sines and Co­sines of the Angles between the wall and the Sun, assuming CB to be the Sine of the greatest Altitude, make CB in the second figure equal thereto, and perpendicular thereto raise the Line CF, wherein prick down the three shadows CE, CD, & CF, and upon B as a Center with the extent BE describe the Quadrant GA, drawing lines from the three shadowes into B the Center, and the Arks GE, GI, GH, are the three Angles between the wall and the Sun, and the nearest [Page] [Page]

[Page 79] distances from the points E, I, H, being taken to GB, are the Sines thereof, also the nearest distances from those Points to AB, are the Cosines thereof.

Being thus prepared, draw three Lines elsewhere as in the third fi­gure meeting in a Center, and making the like Angles as the shadows did, and let them be produced beyond that Center, and have the same Letters set to them: make CH CG in this Scheme, equal to the nearest distances from H I to CB, and make CL equal to EC, in the former Scheme.

Draw the lines IG, IH produced; and make HP GO equal to nearest distances from I H to BG, and parallel to HP and GO, draw the lines IN, IM, each of them made equal to CB, then draw the Hipo­tenusals MO, NP, meeting with the Bases at K and L, draw the Line KL, and it shall represent the Plains Equinoctial or contingent Line.

From I, let fall the perpendicular IQ produced, and it shall be the Substilar line on the Plain and the Point Q is the Equinoctial Point sought.

To place the Substile.

Then repair to the Plain whereon you would make the Dyal re­presented by the first Scheme, and place CQ therein, so that it may make the same Angle with the shadow CE, as it doth with the Line IC in the third Scheme, and it shall be the Substilar Line, which is to be produced, also prick IQ from the third Scheme from C to Q on the Plain in the Substilar Line, and upon the point C, perpendicu­lar to the Substile, raise the Line AC, and make it equal to CB, draw­ing the Line AQ, then will the Angle AQC be equal to the com­plement of the Stiles height; for it is the Angle between the plain and the Equinoctial; just as before in the Horizontal plain, the Sine of the greatest Altitude was erected on a Line falling perpendicular from the foot of the said Sine, to the intersection of the plain of the Suns parallel with the plain of the Horizon, and thereby gave the Angle between the Suns parrallel and the Horizon, equal to the Com­plement of the Latitude.

To find the Center of the Dyal and Stiles height.

If from the point A you raise the Line AV perpendicularly to AQ, where it cuts the Substilar Line, as at V, is the Center of the Dial, and he Line AV represents the Stile.

3. To draw the Meridian Line.

If the Plain recline, a thread and plummet hanging at liberty from the Point B, and touching the Plain, will find a Point therein, suppose K a line drawn from K into the Center of the Dial, shall be the Meridian line of the place.

A broad ruler with a sharp pin at the bottome of it, being in a right line, with a line traced through the length of that ruler, where­to the plummet is to hang, having a hole cut therein for the bullet to play in, will find this Point in a reclining plain.

On an inclining Plain, if the thread and plummet hang upon the Stile at liberty, to some part of the Stile more remote from the Center, fasten another thread, which being extended thence to the Plain just touching the former thread and plummet hanging at liberty, will find many Points upon the Plain, from any of which, if a line be drawn into the Center it shall be the Meridian line required; See the fourth figure: or in either of these cases hold a thread and plummet so at liberty that it may just touch the Stile, and bringing your sight so, as to cast the said thread upon the Center at the same by the interposition of the thread; the eye will project a true Meridian­line on the Plain, for the thread represents the Axis of the Horizon, and the Plain of the Meridian, is in the said Axis.

If the Center fall inconvenient upon the Plain, it will be necessary to draw the Plains perpendicular, passing through the foot of the perpendicular Stile C, and measure the Angle of some of the Sha­dowes from it, and accordingly so place it in the third figure or draught on the floor, on which find the Center, and afterwards as­sign the Center on the Plain where it may happen convenient, and from the said draught, by help of the Plains perpendicular, set off the substile and Stiles height; On plains on which probably the Stile hath but small height, the perpendicular Stile in this work must be assu­med the shorter.

4. To draw the Hour Lines.

These may be set off in a Paralellogram after the Substile, Stile' and Meridian are placed which I have handled before.

Or they may be inscribed by the Circular work, if we assume the perpendicular Stile AC, to be the Sine of the Stiles height, the Ra­dius [Page 81] thereto will be AV, prick the said Radius on the Substile from V the Center to, Z, and upon Z as a Center, describe the Circle as in the first figure, and find the regulating point ☉ by former directi­ons, over which from O the point where the Meridian of the place cuts the said Circle, lay a ruler, and it finds M, on the opposite side, be­ing the Point from whence the Circle is to be divided into 12. parts, for the houres with subdivisions for the halfes & quarters. Here note that the arch MV, is the inclination of Meridians, the whole Semicir­cle being divided but into 90 ^d, and if that be first given (as hereafter) we may thereby find the Point O, whence the Meridian Line is to be drawn into the Center.

5. A Method of Calculation suited hereto.

This is altogether the same as for finding the Azimuth of the sha­dowes on the Horizontal Plain and as easie, whereby in the third fi­gure, the Angle of the Substile CIQ, must be set off from the shadow ICE, just as the Meridian Line CP, might be set off from the shaddow CK on the Horizontal.

And the complement of the Stiles height AQV, in the first fi­gure here, as the complement of the Latitude KTV, was found there.

Lastly, for placing the Meridian Line of the place by calculation, the Substiles distance from the plains perpendicular, and the reclina­tion of the Plain must be found, which may be got easily at any time without dependence on the Sun, and then in the often-mentioned oblique Triangle in the Sphere, we have two sides with the Angle comprehended given, to wit, the complement of the Stiles height, the complement of the reclination, and the Substiles distance from the Plaines perpendicular whereby may be found the inclination of Me­ridians: and consequently the Meridian line of the place, also the Latitude thereof, with the Plains declination, if they be required.

To perform all this other wise Geometrically and instrumentally, (but not by calculation) there was an entire considerable quarto treatise, with many excellent Prints from brass-Plates thereto be­longing, printed at Paris in France, in An. 1643. by Monsieur De­sargues of Lions, which Treatise I have seen but not perused, a year after was published the small Treatise of Monsieur Vaulezard before mentioned.

[Page 82] The inscribing of the Signes, Azimuths, parallels of the longest day, &c. are lately handled by Mr. Leybourn in his Appendix to Mr. Stir­rups Dyalling, as also by Mr. Gibson in his Algebra, who thinks in many cases that they deform a plain, and are seldome understood by the vulgar, wherefore it will not be necessary to treat thereof.

The directions throughout this Book are suited to the Northern Hemisphere and are the same in the Southern Hemisphere, if the words South for North, and North for South be mutually changed.