 The First Book of the Cross-Staff

CHAPT. X. The use of the Line of versed Sines.

This Line of versed Sines is no necessary Line. For all Triangles, both right lined and sperical may be resolved by the three former Lines of Numbers, Sines, and Tangents; yet I thought good to put it on the Staff for more easie finding of an Angle having three sides, or a side having three Angles of a sperical Triangle given. Suppose the three sides to be, one of them 100 gr. the other 78 gr. and the third 38 gr. 30 m. and let it be required to find the Angle, whose Base is 110 gr.

I first add them together, and from half the sum subtract the Base, noting the difference after this manner.

 The Base 110 gr. 0 m. The one side 78 0 The other side 38 30 The sum of all three 226 30 The half sum 113 15 The difference 3 15

For so the proportion will hold.

1. As the Radius to the Sine of one of the sides:
So the Sine of the other side, to a fourth Sine.
2. As this fourth Sine, to the Sine of the half sum:
So the Sine of the difference to a seventh Sine
3. The mean proportional between this seventh Sine and the Radius will shew the Sine of the Complement of half the Angle required.

This done, I come to the Staff, and extend the Compasses from the Sine of 90 gr. to the Sine of 78 gr. which is one of the sides; and applying this extent from the Sine of the other side 38 gr. 30 m. I find it to reach to a fourth Sine of about 37 gr. 30 m. From this forth Sine of 37 gr. 30 m. I extend the Compasses again, to the Sine of the half sum 113 gr. 15 m. (which is all one with then Sine of 66 gr. 45 m.) and this second extent will reach from the Sine of the difference 3 gr. 15 m. to the Sine of 4 gr. 54 m.

Then to find the mean proportional Sine between this seventh Sine of 4 gr. 54 m. and the Sine of 90 gr. I might divide the space between them into two euqal parts, and so I should find the Compasses to stay at 17 gr. whose Complement is 73 gr. and the double of 73 gr. is 146 gr. the Angle opposite to 110 gr. which was required.

But because this division is somewhat troublesome I have therefore added the Line of Versed Sines, that having found the seventh Sine you might look over against it, and there find the Angle. And so in this example having found the seventh Sine to be 4 gr. 54 m. over against this Sine you shall find 146 gr. in the Line of Versed Sines for the Angle required as before.

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