The geometrical square, with the use thereof in plain and spherical trigonometrie chiefly intended for the more easie finding of the hour and azimuth / by Samuel Foster ...

¶By having the Legs and Base, to find the Vertical Angle.

THe Angle given or sought is the Vertical Angle, The sides comprehending it are the legs. The side subtend∣ing it is the Base.

From the top of the Square, count the sum of the legs upon one side, the difference of them on the other side, To this sum and difference apply the threed, Then from the same top of the Square count the base also, And mark where it cuts the threed, for the line passing through the intersection, and standing Square to the top, (if it be numbred from that side of the Square whereon the difference of the legs was counted) gives the Vertical Angle required.

This is the general manner of work for this Proposition, which may be illustrated by these particulars.

FIRST,

Having the Latitude of the place, the Declination and Altitude of the Sun, to find the Hour of the day.

BY the declination of the Sun, may be had his distance from the elevated Pole, By subtracting it from 90 gr. when the Declination is of the same denomination with the said

Pole; Or by adding the Declination to 90 gr. when the De∣clination and elevated Pole are of several denominations.

In this case, we have the three sides of a Spherical Triangle given, and an Angle sought.

The two legs are The complement of the Latitude, and The Suns distance from the Pole. The base is, The complement of the Suns Altitude: The Angle is the Hour required, which must be accounted from the Coast of a contrary name, to the elevated Pole.

According then to the former general prescript, and this par∣ticular declaration, For the hour take the sum and difference of the complement of the Latitude, and of the Suns distance from the Pole, and from the top of the Square, upon one side, count the difference, the sum on the other, to these terms apply the threed; Then from the top of the Square also, count the com∣plement of the Suns Altitude, and where it cuts the threed, the line that crosseth it Square in the same point (being reckoned from that side whereon the difference of the legs was counted) gives the hour from the Meridian or noon.

To make it plain by an Example.

In a North Latitude of 52 gr. 30 min. the Sun declining 20 gr. to the North, the Altitude of the Sun being by obser∣vation 43 gr. I would know the Hour of the day. The legs of this Triangle are the complements of the latitude and de∣clination, that is 37 gr. 30 min. and 70 gr. 0 min. The sum of them is 107 gr. 30 min. their difference is 32 gr. 30 min. Then from the top of the Square at D upon the side DG, I reckon this difference 32 gr. 30 min. downward to k. And on the other side of the Square from the top at B, I also count the sum of the legs 107 gr. 30 min. downward to l. To k and l. I apply the threed. Which done from the top of the Square, again, I count the base 47 gr. the complement of 43 gr. the altitude observed, downward also to 0, and the line that there meets me, I follow till it cut the threed, which is at n, and the line that there ariseth Square to it is n r. I say now that nr, if it be counted from the side DC whereon the difference of the legs was counted, shall give 44gr. 8 min. which turned into hours and minutes of an hour, (allowing 15 gr. to an hour; and 15 min. of a degree to one minute of an hour) will make two hours ••nd 56½ min. from

the Meridian or South, And such is the Hour for that Latitude, Altitude, and Declination.

So also, If in the same Latitude and distance of the Sun from the Pole, but in the altitude of 10 gr. I would know The boure of the day. Here because the legs, that is, the com∣plement of the latitude and the distance from the Pole, are the same, therefore the same position of the threed remaines still, I therefore onely reckon the base (as before) which here is 80 gr. from D to p, then I follow the line p, till it cuts the threed at m, and the line there arising is m s, which counted from DC, whereon the difference of the legs was reckoned, shall give 99 gr. 50 min. that is 6 hours 39⅔ min. of an hour from the Meridian or South.

Another Example.

In the same latitude of 52 gr. 30 min. let the declination of the Sun be 20 gr. to the South, where his distance from the elevated Pole is 110 gr. and let the altitude of the Sun be by observation 10 gr. I require the Hour. The legs are 37 gr. 30 min. the complement of the latitude, And 110 gr. the Suns distance from the Pole. The sum of them is 147 gr. 30 min. The difference 72 gr. 30 min. which I count upon the sides of the Square down to u and t; and the base which is 80 gr. the complement of 10 gr. I count also from D to p, then I follow the line p, till it cut the threed at x, and the line there arising is x y, which counted from DC, whereon the difference of the legs was reckoned, shall give 38 grad. 56 min. that is, two hours and almost 36 min. of an houre from the Meridian or South.

Note, That the threed in this situation, shewes on the dia∣meter of the Square (which in this case represents the Hori∣zon) the Semidiurnal and Seminocturnai Arks, for where the threed crosseth the middle line, the line there arising, (counted from that side of the Square, whereon the difference was num∣bred) shewes the Semidiurnal ark, and counted from the other side, shewes the Seminocturnal ark.

Observe also, If you would known the Crepusculum or Twi∣light, the threed is to be placed as before, according to the sum and difference of the legs, and if you allow 18 gr. for the

Crepusculin line (as they usually doe) the base will alway be 108 gr. which in the two first Examples will not touch the threed at all, and therefore in that latitude and parallel of the Sun, the twilight continues all night. But in the last Example you shall find the Crepusculin line to cut the threed, 6 hours and 15 min. from the Meridian, which shewes that the twi∣light begins at 5¾ a clock in the morning, and ends at 6¼ in the evening, and the rest of the time is dark night which is 11½ hours.

If the sum of the the legs be more then 180 gr. that is, if it would reach beyond the bottom of the Square, you must when you have reckoned to the bottom, count upward back again till you have ended the whole sum.

SECONDLY,

Having the Latitude of the place, the Declination and Altitude of the Sun, To find the Azimuth of the Sun.

HEre also the 3 sides are given, the same with the former, and an Angle sought. The two legs are the Comple∣ments of the latitude, and Suns altitude, The base is the Suns distance from the Pole which is elevated above the Horizon. The angle sought is the Suns Azimuth, from that part of the Meridian, which is of the same denomination with the elevated Pole.

So then according to the former general prescript, and this particular declaration, for the Azimuth, doe thus.

Take the sum and difference of the Complements of the lati∣tude and Suns altitude, and count from the top of the Square, the one upon one side, the other on the other side, and to these terms apply the threed; Then from the top of the Square also, count the Suns distance from the Pole, and where it doth crosse the threed, the line that there ariseth Square to the former, being reckoned from that side of the Square whereon the difference of the legs was counted, gives the Azimuth from that part of the Meridian which is of the same denomination with the elevated Pole, and counted from the other side, gives the Azimuth from the other coast.

To Illustrate it by an Example.

In a North latitude of 52 gr. 30 min. let the altitude of the Sun be 22 gr. and the declination 10 gr. Northerly, By these given I would know the Suns Azimuth, the two legs of the Triangle are the complements of the latitude and Suns altitude, that is 37 gr. 30 min. and 68 gr. the sum of them is 105 gr. 30 min. the difference is 30 gr. 30 min. The sum of them I count on the side BA, from the top at B down to k, The difference I count on the other side from D down to h, and to these points k and h, I apply the threed k h, And last∣ly, because the de∣clination is 10 gr. North-ward in a North latitude, ther∣fore his distance from the elevated Pole is 80 gr. which I count from the top D, down to l, and follow the line at l, till it meet with the threed at n, where I find the line m n, to crosse it also, which numbred from the side DC, whereon the difference of the legs was numbred, gives 102 gr 38 m. the Azimuth from the North: And so also if it be account∣ed from the side BA, it gives the Azimuth from the South 77 gr. 22 min. the residue of the former, or the complement of it to 180 gr.

[illustration]

Another Example, In the same latitude and the same al∣titude, and therefore also the same situation, of the threed, let the declination be Northerly 23½ gr. therefore the distance from the Pole will be 66½ gr. which I count from D to f, and following the line f till it meet with the threed at i, I find the line g i, to crosse there also, which being counted from the side DC, whereon the difference of the legs was counted,

shewes 79 gr. 38 min. the Azimuth from the North, Or count∣ed from the other side, gives the residue of the former, 100 gr. 22 min. The Azimuth from the South.

A third Example. In the same latitude and altitude, and therefore also in the same situation of the threed, let the de∣clination of the Sun be 10 gr. to the South, then shall his di∣stance from the elevated North Pole be 100 gr. and because this 100 gr. is the base, I therefore count it from the top D, down to o, and following the line o, I find it to cut the threed at r, and the line r p there crossing, shewes me from DC, (the side whereon the difference of the legs was counted) 146 gr. 32 min. for the Azimuth from the North, or if the same line be numbred from the side BA, it shewes 33 gr. 28 min. the residue of the former, for the Azimuth from the South.

These Examples may suffice for this kind, and according to these patternes, all others are to be framed.