# Thread: How many pounds + volume of ground beef would fit into an American football?

1. ## How many pounds + volume of ground beef would fit into an American football?

Hi all, I have a bit of a strange question. I'm making an American football made entirely out of meat. (The "Meat Ball") I've already decided to use a bacon weave to wrap it, (for protection, plus to call it a pigskin) but I need to come up with an accurate amount of 1. Volume and 2. Weight so I can fill it with delicious, delicious beef. Here's what I know:

- The ball is 11-11.5 inches long.
- The long circumference is 28-28.5 inches.
- The short circumference is 20.75-21.25.
- There are about 2 cups (1 pint) of ground beef per pound.

This is an entirely serious inquiry. As evidence that I actually make food abominations, here's my previous meat creation (The Bacon Pizza Burger): http://i.imgur.com/nBWbj.jpg

I have one other question to correlate with it: Given a 9x9 inch bacon weave (consisting of 16 slices per weave), how many bacon weaves, and how many slices of bacon, would I need to successfully cover the entire thing? (With some room for the bacon to shrink in the oven as the fat cooks off)

2. ## Re: How many pounds + volume of ground beef would fit into an American football?

Hey klosterdev.

For the weight you simply multiply the density by the volume and then multiply by the gravitational constant. This assumes that everything inside the football is the same thing (i.e. all meat or something else).

If you want to measure the density of meat, then what you need to do is get some standard container with a known volume (in cubic metres) and then get some scales and weigh the meat (i.e. what is on the scales minus the weight of the container), and then take that measurement on the scales (the weight in kilograms) and then divide this by 9.8 and then take this answer and divide it by the volume (in cubic metres which will be really small for a small container).

Now you need to find the volume of your football.

You do this by calculating an integral. I looked at a picture on Google and if the football is mostly circular at every cross section (i.e. if you sliced the football vertically then the cross section of the surface would look roughly like a circle) then you can use calculus to get the volume.

What you do is to fit a parabola and then use a formula in calculus to get the volume and you use this to get the weight.

I'll wait for your response before I go further.

3. ## Re: How many pounds + volume of ground beef would fit into an American football?

From what I remember (Haven't done Calc in a while) the best way to find the volume of a 3d object is to find the equation for the curve then "rotate" it around an access. I'm going to go out on a limb and guess that rotating the long line around would be the more sensible choice?

I guess I have three questions then, in that aspect. First, and this I don't remember at all, what's a good way to find the equation for a parabola using the information that I already have?

Second, and I just want to make sure I have this correct, (I apologize if I format this incorrectly) to find the volume of the football, I try and integrate 2*pi* int[0->14]((f(x))^2 dx ?

Finally, when rotating and integrating a parabola, do I need to do anything special with the equation or is it a pretty basic integration? (I'm rusty on more than the basics at this point)

I appreciate the help.

4. ## Re: How many pounds + volume of ground beef would fit into an American football?

So the best way to think finding the equation of a parabola is that 1) the parabola is symmetric around it's maximum (so basically it has a maximum above the x-axis and then it decreases from both sides and 2) it has roots a and b where f(a) and f(b) = 0.

So what we do to find the equation of the parabola is the following: since the parabola is symmetric it means the roots will be -r and +r for some r and that value of r is half the total length of the football (as if you measured it from tip to tip by cutting it in half and then placing a string from tip to tip and measuring the length of the string).

So if you know this tip to tip diameter, then r is half the diameter.

Know for the maximum. The maximum is where the parabola has it's maximum and by calculus we know that the maximum is when f'(x) = 0 (i.e. the first derivative).

So for a parabola f(x) = ax^2 + bx + c and we have to figure out a, b and c.

Now since the parabola is symmetric around the y-axis we know that the maximum is at x = 0. Now you will know the maximum of the parabola (basically go to the centre of the football along the diameter and measure the radius of that circle-slice cross section).

So we have c = f(0) = this cross-section maximum and we also know that f(-r) = f(r) = 0. So let's solve for a and b.

a*(r)^2 + b*r = -c
a*(-r)^2 + b*(-r) = -c

b*r = b*(-r) which means
b = 0 and now find a where
a*r^2 = -c so
a = -c/r^2 where r is our half-length of the long axis of the football and c is the half-length of the other axis (at the maximum cross section).

So that's your parabola.

Now to get the volume you need to use the volume method which is

Integral from -r to r of pi*[f(x)]^2 dx where f(x) = ax^2 + bx + c.

Now (ax^2 + bx + c)^2 is just a polynomial which means it's easy to integrate.

So that gives you the volume and we have discussed the mass above and after that you're done.

5. ## Re: How many pounds + volume of ground beef would fit into an American football?

For the volume just submerge a fully inflated ball into water and measure the amount of water displaced.
Also, this is one of the best applications of mathematics I've ever seen! Good luck with the abomination.

6. ## Re: How many pounds + volume of ground beef would fit into an American football?

I love it how the simplest ideas always are the ones that really are the best. Thanks pomp!