# Math Help - long compass (arc segment drawing)

1. ## long compass (arc segment drawing)

Hello
i was wondering if you geometry gurus can help me out

if you look at this
Long compass

can you tell me if there is name for this
and can you also help me to formulate a formula to allow me to generate the radius from the angle of the two sticks and the distance of the pins

i am hoping to use this type of movement to cut large radius in plywood
say from 8ft to 75 ft radius over the long side of a sheet of ply

we build concrete skateparks and use large radiuses all over the place

thanks
Bryant

2. ## long compass

posted by junkdog

I have never heard of this but in reading the description it is a method of drawing arcs of large dia circles given three points on the arc.A large arc can also be drawn using a long string of desired radius. Given an arc of unknown radius mark three points equidistant apart and connect the points. Then construct the perpendicular bisectors of each chord. these two lines meet at the center of the circle. Measure radius. Or if the angle between the chords can be determined then the radius can be calculated

bj

3. Originally Posted by junkdog
Hello
i was wondering if you geometry gurus can help me out

if you look at this
Long compass

can you tell me if there is name for this

...
Have a look here: Circle - Wikipedia, the free encyclopedia and then scroll down the page until inscribed angles.

4. Originally Posted by junkdog
Hello
...
and can you also help me to formulate a formula to allow me to generate the radius from the angle of the two sticks and the distance of the pins

...
I've attached a sketch of the situation: The 2 sticks are drawn in green, the angle formed by the 2 sticks is called $\alpha$, the distance between the pins is d and the radius is called r and is drawn in red.

The perpendicular bisector of d, half of d and r form a right triangle.

Since $\sin(\tau)=\sin(180^\circ - \tau)$ you'll get

$\dfrac{\dfrac12 d}{r} = \sin(\alpha)~\implies~\boxed{r = \dfrac{\dfrac12 d}{\sin(\alpha)}}$

5. thats exactly what im looking for

if i rearrange your formula to

angle = asin*(1/2d)/r

would that be correct according to "math rules"
this works for me if i do the calculations im just not sure its correct
been a long time in doing this kind of stuff for me

this gives me a couple of ways of checking my geometry in a spread sheet so i can see if my machine will fit inside my shop as well possible work in a portable unit

also what program did you use to create that formula and nice line drawings from not saying the program did the work it just looks nice

Bryant

6. Originally Posted by junkdog
thats exactly what im looking for

if i rearrange your formula to

angle = asin((1/2d)/r) You have to use brackets!

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also what program did you use to create that formula

have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html

and nice line drawings from not saying the program did the work it just looks nice

Bryant

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