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6 answers

the area would be A =(0.5)*(pi)*r*r, where pi is approximately 3.141592654 and r is the radius of the widow, in this case 7.5 inches. This would give you 88.357 square inches of glass.
Hope this helps.

2006-12-02 15:28:24 · answer #1 · answered by st 1 · 0 0

when you say make, do you mean how large a rectangle of glass will you need to cut a semicircle from? if so, get a square piece:
15 inches on a side which will be 225 square inches.

if you just want the area of the semicircle, it's just half the area of a circle with diameter 15 in.:

A = Pi*(D/2)^2/2 = Pi*D^2/8 = 88.4 sq. in.

2006-12-02 12:33:52 · answer #2 · answered by Anonymous · 1 0

All of you who simply calculated the area are mistaken; you are ignoring the fact that you have to start with a rectangular piece (this is how glass is supplied) and cut it to semi-circular.

If you are cutting on a waterjet, you don't want to fixture the glass, so you need some "extra" to account for variations of how the operator might lay it in the machine. We use 1" extra material in our place for this margin of error.

So, I would say that you want to use an 8" x 16" piece of glass, layout and cut your semi-circle, and know that you are going to toss the rest. 8 X 16 is 128 square inches.

2006-12-03 06:42:37 · answer #3 · answered by www.HaysEngineering.com 4 · 0 0

You are referring to a half moon window I take it?
I calculate 88.3 sq. inches. The 15 inches - does that include the lip to fit into the frame? If not, then you would a little more for that.

2006-12-02 12:28:05 · answer #4 · answered by amland1 2 · 0 0

You need 88.6 in2. For future reference, the "area" of a circle is calculated by radius x radius x 3.1416. Radius is half the diameter. Circumference of a circle is 3.1416 x diameter.

2006-12-02 13:22:25 · answer #5 · answered by questionable reality 3 · 0 0

Formula for area of circle pi*r^2.

r = d/2 = 7.5 in

half of the area is what you need for a semicircle

Area needed = pi * 7.5^2 /2 = ~ 88.36 sq in

2006-12-02 12:30:11 · answer #6 · answered by Anonymous · 0 0

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