# Math: measuring vs: calculating

Here’s a poser. 2 concentric circles, radii of 5" and 5.25". Offset the centers by 2.5". What is the area of the “crescent” formed. This comes to mind 'cause of a project to determine the flow through a gate valve opened a specific # of turns on an acme thread. The only measurements I took were the diameters. the other known value was the specs on the acme thread. It took me a long time, and I.m not sure I was accurate. Any thoughts on the formulas to use? Please don’t be concerned about pressures, material being transferred etc. Area calculations are the question.

(no signature yet but pi r round, cake r square works for now)

Sort of applicable: here.

Give one circle radius r and place it at the origin. It is described by
(1) x[sup]2[/sup] + y[sup]2[/sup] = r[sup]2[/sup]

Give the other radius R and place it a distance d away from the origin, along the x axis. This circle is described by
(2) (x-d)[sup]2[/sup] + y[sup]2[/sup] = R[sup]2[/sup]

Find the value of x where the two circles intersect by inserting (1) into (2); assuming my algebra is correct, this gives
x[sub]int[/sub] = (d[sup]2[/sup]+r[sup]2[/sup]-R[sup]2[/sup])/(2d)

How does this help us? Well, we can solve (1) for y (y=sqrt(r[sup]2[/sup]-x[sup]2[/sup])) and integrate from x = x[sub]int[/sub] to r. Similarly, we can solve (2) for y (y=sqrt(R[sup]2[/sup]-(x-d)[sup]2[/sup])) and integrate from x = R-d to x[sub]int[/sub]

Adding these two integrals and doubling them gives the area of the intersection of the two circles. And then the rest is easy; take the area of one circle and subtract the area of intersection to get the area of that circle that doesn’t intersect the other circle.

Evaluation of the integrals is left as an exercise for the reader, because I don’t integrate at 2 in the morning.

After some searching, I wasn’t able to find a general equation for this, so I’ll have to fall back on calculus.

Ok, center the 5.25" circle at the origin, and the 5" circle with its center shifted to the right at (2.5 , 0 ). This makes the graphs of the two upper semicircles:

• y = sqrt( 5.25[sup]2[/sup] - x[sup]2[/sup] )
• y = sqrt( 5[sup]2[/sup] - ( x - 2.5 )[sup]2[/sup] )

Set them equal to each other to find the intersection:
27.5625 - x[sup]2[/sup] = 25 - x[sup]2[/sup] + 5x - 6.25
5x = 27.5625 - 25 + 6.25 = 8.8125
x = 1.7625

Now find the area under the larger semicircle by integrating
sqrt( 5.25[sup]2[/sup] - x[sup]2[/sup] )
for x in [-5.25, -2.5] (range not covered by the small circle).

Then, for x in [-2.5, 1.7625], find the area under the larger circle and above the smaller:
sqrt( 5.25[sup]2[/sup] - x [sup]2[/sup] ) - sqrt( 5[sup]2[/sup] - ( x - 2.5 )[sup]2[/sup] )

Add the two areas together, then double it (since the above was only dealing with the upper semicircles). Unfortunately, my memory of the trig substitutions neccesary to integrate sqrt( a[sup]2[/sup] - x[sup]2[/sup] ) have been replaced by Buffy episode descriptions and video game cheat codes since college Oh, and I’m assuming you wanted the larger crescent (on the left in the way I set up the graph). If you wanted the crescent with the outer edge formed by the smaller circle, find the areas under:
sqrt( 5[sup]2[/sup] - ( x - 2.5 )[sup]2[/sup] ) - sqrt( 5.25[sup]2[/sup] - x [sup]2[/sup] ) from [1.7625 , 5.25 ]
and
sqrt( 5[sup]2[/sup] - ( x - 2.5 )[sup]2[/sup] ) from [5.25 , 7.5]

On preview, I see g8rguy with his little namby-pamby explanation with x’s, y’s, and r’s, and it’s functionally the same as mine (including leaving the actual integration up to the student). But I typed all this out, damn it, I’m posting!

Thanks guy’s and Q.E.D. Wow, that’s a lot more than my trig and intro to calc can interpret tonight, but you’ve answered as I requested. I’ll digest it over the next few days. The original problem was to cut the bottom of a cone attached to the bottom of the gate valve so that with the valve fully opened the flow would be limited by the cone to the 12 turns on the ACME thread.
I was happy for the 3 days figgering it out, and the dummies were happy with the cone I installed! Actually they didn’t know the flow rate they started with. It was steeped barley from a conical tank bottom into a conveyor to be malted for BREWSKY!

Minor correction: when I say integrate from R-d to x[sub]int[/sub], I really mean d-R, actually. Embarrassing, but hey, it was late.

The actual integration isn’t too bad; sqrt(r[sup]2[/sup]-x[sup]2[/sup]) can be done by writing x = r * cos(q), which gives -r[sup]2[/sup] sin(q)[sup]2[/sup], integrated on q. And that can be either looked up in a book or, if you’re feeling daring, can be done by writing 2 * sin(q)[sup]2[/sup] = 1 - cos(2q).