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-   -   Crazy Math Formulas (http://www.chiefdelphi.com/forums/showthread.php?t=20009)

patrickrd 19-09-2003 00:12

Quote:

Originally posted by Scott England
-d(a(du/dx))/dx+(d^2(b*(du^2/dx^2)/dx^2))+co*u+c1*(du/dt)+c2*(du^2/dt^2)=f(x,t)

this equation represents unsteady heat transfer in a fin

Actually, I think the c1*(du/dt) term is irrelevant. c1 will always be zero because the derivative is taken twice (with no intermediary steps in between) of the governing law (which is a function of t^2 only, not t), thus integration is guaranteed that the constant must equal zero. For comparison, consider an example of a sinusoid that is translated vertically in the y direction:

f(t) = sin(t) + 2

Then:

f'(t) = cos(t)
f''(t) = -sin(t)

Integrating twice yields:
f''(t) = sin(t) + C1t + C2

However, we are guaranteed that the sinusoid is translated, and should have no t term, thus C1 MUST be zero. With your above equation it's the same idea. c1 must be zero.

Seriously, putting aside my lame and incorrect argument above that has no validity at all, heat transfer equations (specifically fin analysis) are without any doubt in my mind the absolute worst equations I have ever come across. I'm actually impressed you found an equation for fins that is so simple... probably because it's in it's most generic form. Anyway, if you really like I can post the most obscene equation I can find in my heat transfer book. I remember it being very long, and containing many variables that took other equations to define what those variables mean, and so on and so on several times.

- Patrick

Yan Wang 19-09-2003 18:36

Lol, well Pat, according to a materials engineer like my dad, those "aren't hard at are to know by heart." Pfft. :P

George1902 19-09-2003 21:47

I don't know about crazy, but I've always enjoyed this:

Quote:

(12+144+20+3*(4)^1/2)/7 +5*11 = 9^2 + 0

That is:

A dozen, a gross, and a score,
Plus three times the square root of four,
Divided by seven,
Plus five times eleven,
Equals nine squared and not a bit more

-John Saxon
George

echos 30-12-2003 18:39

Re: Crazy Math Formulas
 
Not much of a challenge but definatly a time filler.

pi/2 = 2/1 + 2/3 + 4/3 + 4/5 + 6/5 + 6/7 + 8/7 + 8/9 + 10/9...

It just repeats on and on.

Adam Y. 02-01-2004 12:20

Re: Crazy Math Formulas
 
Quote:

-d(a(du/dx))/dx+(d^2(b*(du^2/dx^2)/dx^2))+co*u+c1*(du/dt)+c2*(du^2/dt^2)=f(x,t)
This reminds me of an equation that my friend showed me that related to trebuceuts. The thing was a page long differential equation.

swando 07-01-2004 15:43

Re: Crazy Math Formulas
 
Quote:

Originally Posted by echos
Not much of a challenge but definatly a time filler.

pi/2 = 2/1 + 2/3 + 4/3 + 4/5 + 6/5 + 6/7 + 8/7 + 8/9 + 10/9...

It just repeats on and on.

i dont get it...
pi/2 is ~1.57, the first number on the other side of the = is a 2. and you keep adding more?
hm..

IMDWalrus 07-01-2004 15:50

Re: Crazy Math Formulas
 
Quote:

Originally Posted by echos
Not much of a challenge but definatly a time filler.

pi/2 = 2/1 + 2/3 + 4/3 + 4/5 + 6/5 + 6/7 + 8/7 + 8/9 + 10/9...

It just repeats on and on.

How does this one work? As swando said, 1.57 =/= 2. I'm thinking that that pi/2 is supposed to be something else, but I have no idea what...

echos 21-01-2004 17:23

Re: Crazy Math Formulas
 
it give exactly half the value of pi thats all the pi/2 means. it means pi divided by 2 = 2/1 ...

swando 21-01-2004 17:45

Re: Crazy Math Formulas
 
Quote:

Originally Posted by echos
it give exactly half the value of pi thats all the pi/2 means. it means pi divided by 2 = 2/1 ...

but pi/2 > 2/1 + 2/3 already.
after a quick google i found the real formula:
pi/2 = 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7...

echos just made a symbol error :)

echos 22-01-2004 16:51

Re: Crazy Math Formulas
 
Thanks for the correction it was a little fuzzy in my mind. I found that one on google a while back ago and made a c program out of it. but i lost the source code.

Code:

unsigned double float pi;
 unsigned int top;
 unsigned int bottem;
 unsigned int count;
 
 int main()
 {
 top = 2;
 bottem = 1;
 pi = top/bottem;
 
 while(count < 10000)
 {
 pi *= top/bottem;
 bottem += 2;
 
 pi *= top/bottem;
 top += 2;
 
 count++
 }
 
 printf ("pi = %f", pi*2);
 }


Grommit 08-04-2004 23:47

Re: Crazy Math Formulas
 
Try solving problem #2 in this thread:

http://www.chiefdelphi.com/forums/sh...ad.php?t=27647

Aside from that, I recommend looking at Olympiad problem books with inequalities in them, you can get some crazy formulas in there.

Talk to me on AIM or send me a PM if you want me to suggest some good books... I know quite a few.


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