PROBL. VII. How to Measure a Conical piece of Timber.
ADmit you had a Cone Piece of timber whose Base or Diameter at the End AB is 28 inches, and the length thereof CD 15 foot, it is required to know how many feet of timber is in the Piece.
Extend the Compasses always from 23 43/100 unto the Diameter AB 28, the same distance will reach from the length 15 foot turned twice over unto 21 4/19.
ABq × CD / 550 = 21 38/100 foot of timber in the Cone Piece.
And by the former Rule you may Measure any part of a Cone or Pyramide-piece. Admit you were to cut a Piece of 5 foot at the greater End, and you find the Diameter EF 18:95 Inch. First, Mean-diameter 18:95 and 28 Inch. added is 46; 95 the half is 23:48 the Mean: then extend the Compasses from 13 54/100 unto the Mean-dia∣meter 23 48/100; the same distance twice repeated from the length 5 foot, will reach to 15 ••7/10 foot in the ⅓ of the Cone at the great End; And likewise to Measure EFHG the Diameter HG is 9. 45 added to 18 95 EF, the Sum is 28:4.
The ½ is 14; 20 Inches the Mean-proportion the length 5 foot; by the former Rule you will find in that Piece of timber 5 50/100 foot; and to Measure the little Cone GH 9 45/100 inches diameter and 5 foot long; Work as to Measure the whole Cone, and you will find it 81/1••0 parts of a foot.
| foot. | parts. |
| 15 | 07 |
| 5 | 50 |
| 0 | 81 |
| 21 | 38 |
And so you have truly Measured the Pieces, as you may find by adding them up, and they make 21 foot 38/199 parts, as you found in the whole Cone at first; and so by finding the Area of the Circle and •• part, you may find the Segment of any Cone or Pyramide that is Square in the sides by the Area thereof; by the same Rules you Measure Stone. It is needless to make more Examples in this thing.