Geometry Probem

Hi Jim. Given the length y of the circle's chord, and x of the sagitta, the radius of the circle can be calculated as:

r = y^2/8x + x/2

See:

And the height of the sagitta x can be calculated from the radius and the length of the chord y via:

x = r - sqrt(r^2 - (y/2)^2)

Two equations, two unknowns x and r, so it should be possible to solve for r directly.

There may be an easier way to do it, but that's the first that came to mind.

Look at my drawing again. I know only "H", not the dimension ("sagitta") above chord labeled "W".

That's the snag ;-) ...Jim Thompson

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

"H" is not a common trig function, but is related to a sagitta

A practical solution is to temporarily assign radius = 1 and chord "X" =1, or R=X Get a piece of paper and draw it. The angle becomes 60 degrees. Then fit the second chord "W" using "H"

It's OK, you don't have to know it. You know the chord, and you have two equations involving the chord, the sagitta, and and the radius. Two equations, two unknowns, solve them and you get the radius plus the height of the sagitta for "free." The downside is you probably have to solve the two equations numerically. I've asked some of my more math inclined friends if there's an easier, geometric way to do it.

That's what I have been in circular orbit over. I thought you'd have a closed form solution, albeit simultaneous equations >:-] ...Jim Thompson

Arg! I made a dumb mistake. If I'd looked carefully, the two equations I'm using are actually the same equation! Sorry, I thought there was an easy solution, but I've just managed to reveal my own ignorance!

It's subtlely difficult. I keep coming up with redundant equations :-( ...Jim Thompson

I think I see a geometric solution now, but it may be one you've already tried. I'll get back in a bit.

Try this, with a being the sagitta of chord X, and b being the sagitta of chord W. Unknowns a, b, R, D.

a = R - sqrt(R^2 - (X/2)^2)

b = R + D - sqrt((R+D)^2 - (W/2)^2)

(R - a)^2 + (X/2)^2 = R^2

(R + D - b)^2 + (W/2)^2 = (R+D)^2

Looks like these equations may be redundant. :(

I think it's pretty easy, given the hints so far and the right approach.

If you combine the sagitta equation with the two proportional triangles you should have a closed form solution without simultaneous equations.

It's easy to solve for R from the known variables and the first sagitta H .. R = X^2/(8*H) + H/2

To get D, consider the second saggita, call it H'.. we can see that

(D + R - H)/R = (H - H')/D (all of which are known except H')

Simply solve for H' and put it in the sagitta equation, and out pops R+D, and R is known.

Best regards, Spehro Pefhany

Le Sun, 01 Sep 2013 15:18:53 -0700, Jim Thompson a écrit:

Ahem, that's dead easy!

Say alpha is half the sector angle and x = X/2 w = W/2

Then Sin(alpha)=x/r Sin(alpha)=w/(r+d) H=x+r(1-Cos(alpha))

Then, when solving you find *two* solutions:

r1=(H w x-Sqrt[x^4 (H^2+w^2-x^2)])/(w^2-x^2) d1=(H w x-Sqrt[x^4 (H^2+w^2-x^2)])/(w x+x^2) or r2=(H w x+Sqrt[x^4 (H^2+w^2-x^2)])/(w^2-x^2) d2=(H w x+Sqrt[x^4 (H^2+w^2-x^2)])/(w x+x^2)

Of course, so that the student has to do his homework, some errors slipped into the given soltions.

Exercise left to the student to correct them.

I understand which triangle the proportion (D + R - H)/R is referring to (the one bordered by R-X-R, but I don't understand which triangle (H - H')/D is in reference to for the proportion.

Not everyone has access to binaries, maybe put the pdf on your website.

Art

If the sagitta is small compared to the radius, the approximation s = l^2/2r can be used. Then the above equations can be solved for R and D without having to use numerical methods, and I get (using Mathematica):

D = sqrt(2)/4*(sqrt(2H^2 - WX + X^2) + 2H)

R = (sqrt(2)X)/(4(W-X))*[sqrt(2H^2 - WX + X^2) + 2H]

Best regards, Spehro Pefhany

Got it, that's clever. Thanks!