are why I took one semester on the engineering path in college and decided that maybe it wasn't for me.
Random Question that Relates to the Big House
Instead of the stadium itself?
you'd just need to find the dimensions and calculate the cubic volume like it was a giant cylinder drinking glass.
I don't know the answer, nor am I particularly interested in figuring it out.
wouldn't it leak out the exits?
Is this inclusive of the suites?
Either way, I would say it's more than my annual beer consumption by at least a 12er
- All exits are plugged so we're talking to overflow the sides.
- The shape is an idealized cone (cap removed) open up to the sky with the slope of the sides at a 41 degree slope.
- Adjusting for the water seepage into the concrete being negligible
- The long side to be approximately 130 yards long by 70 yards wide at the base. The top adding and extra 4 feet per row, rounding to 100 rows. That's and extra 800 feet added to the base, so ~350 yards by 250 yards.
- Height could be guessed at 8 feet of playing field, plus 3 feet per row, plus the top row having a 5 foot wall behind it, so ~315 feet.
Adjusting for wind, I'd say a lot. Perhaps slightly less at high tide.
And yes, I realize those dimensions are almost decidedly ridiculous. If you're pointing that out, you missed the point of my answer.
the exact dimensions?
Mgoblue has dimensions for Crisler but not Michigan Stadium. As this isn't a scientific study, I might suggest google mapping it from above for width. Height is still a tough call. My joke estimate was WAY ridiculously tall. I could see it being somewhere closer to 50-75 feet, but even that just seems high. Damn bowl in the ground effect messing with my perception.
my kind of math!
Who thinks of this? You may have a little too much time on your hands.
Honestly, I once wanted to know how much beer could be fit into the stadium. But I was thinking stacked kegs. The volume is pretty insane.
This is a very quick and dirty back of the envelope calculation, and working with FA assumptions of a height of 300 feet and size of 350 yards by 250 yards (or 1050 x 750 feet). For easy of math I assumed a 45 degree angle for the seats...
The volume of the field would be 100 yards x 50 yards x 300 feet or 300 feet x 150 feet x 300 feet for a volume of 13,500,000 feet^3
The volume of the sides ends up being 300 feet x 300 feet x 300 feet for the long side, 300 feet x 150 feet x 375 feet for the short side and 300 feet x 300 feet x 375 feet for the corners (again assume 45 degree angles so each side volume is half of a rectangle and the corner is 1/4 of the rectangle)
The total volume from my half-assed work is 91,125,000 feet^3. Convert that into whatever units you like. For gallons is ~681,662,000 gallons.
Excellent work. Having just been informed that I drank a fifth of vodka solo last night, math wasn't about to get done. Also, 315 feet sounds awfully tall. 30 stories? I don't think so. I'm guessing 3 feet tall rows was a bit generous. 2 foot would still have it around 20 stories. Maybe 18 inches, but even 15 stories sounds high.
That's still way too generous. Think about it, are the rows different in height by the difference between the bottom of your foot to roughly your knee/mid thigh?
Thinking about it another way, when you enter the stadium, you're at around row 70. There's 40ish rows above you, and from the exterior, that's what... 40 feet at most?
That means the rest of the rows account for another 70 or so feet, + 5 feet or so to get to the field level. The stadium is somewhere around 115-125 feet from the bottom to the top row, or around 10-11 stories.
Cowboys new stadium is 104 million cubic feet for a rough comparison.
Thanks. It's good to know I'm within the correct order of magnitude at least. Math for the win!
Why are you so long?
ahhh, can't get it to embed...
Blueprints on the wall of the M-Den on State St. Second floor by the front stairs. That should help you answer the question.
Football season cannot get here soon enough...
why are you trying to get us to do your math homework?
This gives me something to distract from my engineering HW (spring semester, how I loathe thee...)
Assumption: Filling to the top of the bowl, the stadium space can be approximated as a "pyramidal frustum" (basically a rectangular pyramid with the top cut off). This neglects the curvature of the seating areas as well as the rounded corners of the seating areas. Since one would add to the overall volume and the other would detract, I don't feel too bad about it.
The volume of a pyramidal frustem is given by:
V = (h/3)[B1+B2+sqrt(B1B2)]
From Google Maps and a diagram from the stadium renovation website (http://www.umich.edu/stadium/project-description/pdf/061128Plans.pdf), I came to these approximate dimensions:
L1 = 400 ft W1 = 220 ft (field level)
L2 = 800 ft W2 = 620 ft (top of bowl)
h = 85 ft
V = (85/3)[88000 + 496000 + 208921] ft^3
V = 2.25 e7 ft^3
That's 22500000 cubic feet
or 636000 cubic meters
or 168 million gallons
Or, approximately the current size of the Deepwater Horizon Oil Spill. Wow, how's that for topical?
Why don't you just use a triple integral to calculate the volume of a oval cone from the top to about 1/3 of the way down? Then you can approximate the angle of decline, and integrate to get a much more precise measurement, especially since all you have to do is measure the two radii.
After that, add in some rectangular prisms for the new luxury boxes and you're all set. I'm not doing it because fuck that noise, I took calc III 3 years ago.
Excellent. Now we know how big the Big House is. Your comment about the Deep Water Horizon Oil Spill got me going. It was actually 200 million gallons, or that is the latest high estimate. So if we make the Gulf of Mexico the size of Michigan Stadium , what is the equivalent size of the oil spill?
The Gulf is 660 x 10^15gallons. 200million is 3.1 x 10^-8% of that. and 3.1 x 10^-8% of 168 million gallons is 7 ounces! The "biggest environmental disaster of all time" is 7 ounces in Michigan Stadium. Not even one beer. Imagine that.
Estimate the area of the field level (square feet). Estimate the are of the top of the bowl (square feet). Add them together. Divide by 2. Multiply by the height of the stadium (feet). Multiply by 7.48 to covert cubic feet to gallons.
I'm sure mathlete is working to get you an answer right now.
How much water can it hold? Come on. That's too easy. How much milk can it hold? Now that's a solid question. somthing I've always wondered...