Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**ninthhour****Member**- Registered: 2005-08-15
- Posts: 1

Please help me figure out the total volume of a quonset hut. I need to know how many cubic feet it can hold. Width=80 height = 24 length= 205 all in feet

I need to know how many cubic feet it will take to fill it completly.

here is a link so you can see what I am talking about.

http://en.wikipedia.org/wiki/Quonset_hut

Thanks for your help

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 12,968

A quonset hut would be half of a right solid cylinder. Length is given as 205 feet. Width is given as 80 feet, and height 24 feet. But if the width is 80 feet, the height would have to be 25.46479 feet. That is obtained by dividing 80 by pi.

The volume would be pi*r²*h/2 = (3.141592 x 648.4556 x 205)/2

= 208,811.285 cubic feet

Character is who you are when no one is looking.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Isn't the width the diameter and the height the radius? And if so, wouldn't height be width/2?

Also, it would probably make more sense if the hut's cross-section was a segment rather than a semi-circle, because then the height and width could both take the given values. Unfortunately working out the volume of a segmentular prism (?) involves more complex maths than before. Whatever the calculated value is, you should probably round it down a bit if you are thinking of using the information practically, because some of the volume will be impractical as storage space, unless you're filling it with gas or something.

Why did the vector cross the road?

It wanted to be normal.

Offline

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

It looks like the Wiki's designation of the quonset as a smeicircle is incorrect--it's not limited to being a half-circle only. Otherwise, the dimensions of the original quonsets that they published would be incorrect. So, it must be a segment of a circle, and we need to find the radius of that circle.

Let's center our circle at the origin and draw two vertical lines at x=40 and x=-40 (so the distance between them is 80). We know that our circle intersects both lines, and that the "top" of the circle (where it intersects the y axis) is 24' higher than that point of intersection. Connect the points of intersection, and you have your secant line.

Now:

r = (c² + 4h²) / (8h)

Source: The Math Forum

c = 80

h = 24

r = (80² + 4*24²) / (8 * 24) = 45 1/3

Now, you need to find the area (K) of the segment, but first, you need to find the central angle (theta). The formulas from MathForum:

theta = 2 arcsin( c / [2r] )

K = r²[ theta - sin(theta) ] / 2

You can then "extrude" your segment area by multiplying by the length of your quonset.

*Last edited by ryos (2005-08-16 12:07:53)*

El que pega primero pega dos veces.

Offline

**John E. Franklin****Guest**

I visited the web site mentioned by the person who started this thread. The photos of a Quonset hut reveal that they are about a third of a circle, not a semi-circle. What I mean is that the center of the circle is below the ground about ten feet.

**John E. Franklin****Guest**

With the center of the circle 16 feet (40-24) below ground, I made a rough estimate of the answer.

I obtained a volume of 260400 cubic feet. This may be off by a percent or two because of the crude method I used.

Pages: **1**