THESE FORUMS NOW CLOSED (read only)
Comic Discussion => QUESTIONABLE CONTENT => Topic started by: oldbushie on 18 Nov 2008, 13:21
-
My mind goes off on tangents just like hers. It's bizarre what I end up thinking about sometimes. XD
-
Ever seen one of those "Guess the number of marbles in the jar, win a prize!" features of a church carnival or whatnot? Whenever I encounter one, I have to sit cross-legged on the grass, whip out a pen, paper, and calculator and proceed to take representative counts of part of the jar and make simplifying assumptions and calculations based on the geometry of the marbles (or pennies or stones or...) and the geometry of the jar and 20 minutes later come up with what I considered to be the upper limit on the count and then make an error analysis to reduce that to come up with my guess.
-
Ever seen one of those "Guess the number of marbles in the jar, win a prize!" features of a church carnival or whatnot? Whenever I encounter one, I have to sit cross-legged on the grass, whip out a pen, paper, and calculator and proceed to take representative counts of part of the jar and make simplifying assumptions and calculations based on the geometry of the marbles (or pennies or stones or...) and the geometry of the jar and 20 minutes later come up with what I considered to be the upper limit on the count and then make an error analysis to reduce that to come up with my guess.
I had a math teacher who made that into a weekly quiz.
-
Ever seen one of those "Guess the number of marbles in the jar, win a prize!" features of a church carnival or whatnot? Whenever I encounter one, I have to sit cross-legged on the grass, whip out a pen, paper, and calculator and proceed to take representative counts of part of the jar and make simplifying assumptions and calculations based on the geometry of the marbles (or pennies or stones or...) and the geometry of the jar and 20 minutes later come up with what I considered to be the upper limit on the count and then make an error analysis to reduce that to come up with my guess.
but do you win?
-
"Writing this actually gave me a headache." It gave me a boner, and i'm waaay too old for boners. Please stop teasing us with Hannelore - there are no real women like that. Or not nearly enough, anyway.
-
Am I the only one who wants Panel 3 as a desktop background? For some reason I just like that crazy fish-eye effect :-D
-
The easiest way to get the correct answer is to average all the other guesses and round. Most likely to be right.
-
You want the correct answer? Cheat. Buy an exact replica of the jar, fill it with your own jelly beans, then replace the real jar with yours.
-
What if they count before putting them in the jar, so they dont have to do it later?
-
There is only one way find the correct number of beans. Using high school level super-mathematics!
(http://superdickery.com/images/stories/oneshot/1007supermathematics1iw.jpg)
Incidentally, math is not Superman's strong point.
-
I created an account just to login and request that Jeph show us the text that is behind her head.
PLEASEFORTHELOVEOFGODIMUSTKNOW.
-
If you go back and read 787, you find a very interesting contrast.
Ever seen one of those "Guess the number of marbles in the jar, win a prize!" features of a church carnival or whatnot? Whenever I encounter one, I have to sit cross-legged on the grass, whip out a pen, paper, and calculator and proceed to take representative counts of part of the jar and make simplifying assumptions and calculations based on the geometry of the marbles (or pennies or stones or...) and the geometry of the jar and 20 minutes later come up with what I considered to be the upper limit on the count and then make an error analysis to reduce that to come up with my guess.
but do you win?
He did give the winner a candy bar.
-
I do math problems in my head just to see what the answer REALLY could be....
....i'm so ashamed.
-
Friggin' math people. :-P
I make History Channel-style documentaries in my head when I play games like Rome Total War....or just for fun.
-
I do math problems in my head just to see what the answer REALLY could be....
....i'm so ashamed.
In class we've been covering the behavior of fluids. We were discussing density and buoyancy. I gave the class a fairly complex problem I knew they didn't have the tools to solve, then began walking them through it. Then I realized how very far beyond them it was and stopped. Then I realized the answer was fairly trivial and started working on a more difficult problem.
The easy problem was:
You have a cube of density ρc<ρf the density of a fluid. If the cube is of side length l, how far into the fluid will it sink?
It quickly becomes apparent that if you state that d is the length of the vertical edge below the surface of the water then d/l = ρc/ρf.
Now, a much more difficult problem is:
You have a right cone of height h, radius R, angle α, and density ρc in a fluid of density ρf, how far will it sink (base first) into the fluid?
You have to be able to find the roots of a cubic equation (it turns out to be surprisingly simple, actually. The equation, that is), but the answer is: d/h = (ρc/ρf - 1) ^ (1/3) + 1
You'll notice that this always gives an answer between 0 and 1, which is what you'd expect. Also, the closer the density of the cone is to the density of the fluid, the greater the depth to which the cone is submerged.
Then I tried to find out how far the cone would sink point first and made a mistake somewhere and came up with an answer such that the denser the cone, the less it's submerged. But I was way, way too tired to find my mistake.
-
Applause for wide angle focus Hannelore. APPLAUSE, I SAY. Bravo, encore, etc.
-
"Writing this actually gave me a headache." It gave me a boner, and i'm waaay too old for boners. Please stop teasing us with Hannelore - there are no real women like that. Or not nearly enough, anyway.
Creepy!
-
Then I tried to find out how far the cone would sink point first and made a mistake somewhere and came up with an answer such that the denser the cone, the less it's submerged. But I was way, way too tired to find my mistake.
That may actually have been right, since the submerged volume would increase more and more rapidly as it sinks. Though obviously at some point it would have to just sink.
Anyway, it doesn't really matter, because you couldn't get a cone to float like that; point first is an unstable equilibrium, so it would quickly flip over and turn into the other problem. :P
-
Actually, it would lay on its side and become a different problem entirely.
Still, I found my mistake. I was thinking of 'd' as the submerged distance, but I'd written it and solved it as the unsubmerged distance. Now my answer makes sense; the denser the cone gets, the less unsubmerged it is.