jon virgi

04-07-2003, 09:09 PM

does anyone know any crazy math formulas or sites for them?

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jon virgi

04-07-2003, 09:09 PM

does anyone know any crazy math formulas or sites for them?

srjjs

04-11-2003, 12:31 AM

Just what do you mean by crazy?

Etbitmydog

04-11-2003, 12:55 AM

If you're going into engineering as a major, you'll come accross this. This is a dynamics formula that relates acceleration of one point, to the acceleration of another point, to the angular acceleration of the object, and angular velocity of the object and a velocity relavie to another point. the parentasies are added because that means that the vector is such relative to such. All the following are vectors.

Since i don't have all the proper symbols,

I'll use % for angular velocity

$ for angular acceleration

a(B)=a(a) + $Xr(B/A) + %X[%Xr(B/A)] + 2%X(v(B/A)+a(B/A)

X stands for cross product

r is a vector and (B/A) means that the vector at point B is relative to A.

This is math.. Isn't it?

Since i don't have all the proper symbols,

I'll use % for angular velocity

$ for angular acceleration

a(B)=a(a) + $Xr(B/A) + %X[%Xr(B/A)] + 2%X(v(B/A)+a(B/A)

X stands for cross product

r is a vector and (B/A) means that the vector at point B is relative to A.

This is math.. Isn't it?

Jim

04-11-2003, 05:08 AM

http://antigravitypower.tripod.com/MathProblems/index.html

jon virgi

04-11-2003, 06:09 AM

I mean either like really tough problems, or even better, really weird, or odd formulas. Ive seen one before that was able to tell the amount of miles a tornado would last based upon its wind speed. Do you know any odd or weird formulas?

Scott England

04-13-2003, 08:58 PM

-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 is tough to type in without an equation editor, (du/dx) denotes the partial derivative of u with respect to x, (du^2/dt^2) denotes the second partial derivative of u with respect to t, etc.

anyways, this is a model equation from my finite element analysis class

When b=0, c2=0, co=0, c1=rho*A, a=kA

this equation represents unsteady heat transfer in a fin

When a=0, b=E*I, co=k, c1=0, c2=rho*A

this equation represents transverse vibrations in a beam

When a=E*A, b=0, c1=0, c2=rho*A, co=0

it represents longitudinal motion(vibration or wave propogation) of a slender rod

all of these require initial conditions and boundary conditions to be specified, but that would have been alot more typing and i think it looks intimidating enough already

this is tough to type in without an equation editor, (du/dx) denotes the partial derivative of u with respect to x, (du^2/dt^2) denotes the second partial derivative of u with respect to t, etc.

anyways, this is a model equation from my finite element analysis class

When b=0, c2=0, co=0, c1=rho*A, a=kA

this equation represents unsteady heat transfer in a fin

When a=0, b=E*I, co=k, c1=0, c2=rho*A

this equation represents transverse vibrations in a beam

When a=E*A, b=0, c1=0, c2=rho*A, co=0

it represents longitudinal motion(vibration or wave propogation) of a slender rod

all of these require initial conditions and boundary conditions to be specified, but that would have been alot more typing and i think it looks intimidating enough already

jon virgi

04-15-2003, 08:08 PM

sweet, does that formula have a name or something, so i could find more out about what it does and such? what is rho*A? Is kA supposed to be a an acid equilibria constant(im obviously thinking chem)? Thanks for the formula.

Scott England

04-15-2003, 09:13 PM

we just call this equation the model equation, since by varying the coefficients you can model tons of stuff, from vibrations in any which direction, heat transfer through conduction or convection, probably lots more stuff but that's the bulk of what we focus on.

Its just a good overall exercise since it includes second and fourth order spatial derivatives and first and second time derivatives.

k can be any number of things, but my major is ESM(engineering science and mechanics), some places call it Applied Mechanics, in either case, i haven't touched most forms of chemistry since freshman year. So no, kA is not a chemical property. In heat transfer problems, k*A would be thermal conductivity of the material*cross sectional area. In solid mechanics and dynamics, k would be a spring constant, or a radius of gyration (neither of those showed up in the examples i typed in)In solid mechanics problems, rho*A would be density of the material times area, E*A would be Young's Modulus * area. Etc.

There's lots more about the Finite Element method than i can ever write in a CD post, i'd highly recommend taking a course in it in college. I believe most companies use it fairly extensively as a more realistic way of modeling things that don't perfectly match text books (hopefully any of the Professional Engineers in here can back me up on that) I just took the Fundamentals of Engineering exam (all 8 hours of it) on saturday, so i've got at least 5 years to go before I can be Scott England, P.E. and now at least 2 years of grad school before i'll ever hit a real company extensively.

Its just a good overall exercise since it includes second and fourth order spatial derivatives and first and second time derivatives.

k can be any number of things, but my major is ESM(engineering science and mechanics), some places call it Applied Mechanics, in either case, i haven't touched most forms of chemistry since freshman year. So no, kA is not a chemical property. In heat transfer problems, k*A would be thermal conductivity of the material*cross sectional area. In solid mechanics and dynamics, k would be a spring constant, or a radius of gyration (neither of those showed up in the examples i typed in)In solid mechanics problems, rho*A would be density of the material times area, E*A would be Young's Modulus * area. Etc.

There's lots more about the Finite Element method than i can ever write in a CD post, i'd highly recommend taking a course in it in college. I believe most companies use it fairly extensively as a more realistic way of modeling things that don't perfectly match text books (hopefully any of the Professional Engineers in here can back me up on that) I just took the Fundamentals of Engineering exam (all 8 hours of it) on saturday, so i've got at least 5 years to go before I can be Scott England, P.E. and now at least 2 years of grad school before i'll ever hit a real company extensively.

srjjs

04-16-2003, 05:37 PM

Einstein+Pythagoras:

E= m c^2= m ( a^2 + b^2)

The limit as n goes to infinity of sin (x) /n is 6.

Cancel the n in the numerator and denominator.

E= m c^2= m ( a^2 + b^2)

The limit as n goes to infinity of sin (x) /n is 6.

Cancel the n in the numerator and denominator.

jon virgi

04-17-2003, 11:30 AM

your einstein + pythagoras does not work. c in the einstein equation is the speed of light (constant )(3x10^8). Einsteins equation can be used to calculate stuff like bond energies and crap like that in MeV (mega electron volts).

now with the limit you posted , Lim n->infinity (sin(x))\n = 0, and not 6. there's no n in numerator, so denominator gets infinitely large, hence, it equals 0.

If you had something like Lim x->0 (sin(x))/x , then the answer is 1. since lim x->0 sinx/x = 0/0 , then we can use L'Hopitals rule, take derivative, get Lim x->0 (cos(x))/1 = 1/1=1

I put that in case u were unsure of limits, b/c the one u gave did not make sense.

now with the limit you posted , Lim n->infinity (sin(x))\n = 0, and not 6. there's no n in numerator, so denominator gets infinitely large, hence, it equals 0.

If you had something like Lim x->0 (sin(x))/x , then the answer is 1. since lim x->0 sinx/x = 0/0 , then we can use L'Hopitals rule, take derivative, get Lim x->0 (cos(x))/1 = 1/1=1

I put that in case u were unsure of limits, b/c the one u gave did not make sense.

srjjs

04-17-2003, 06:15 PM

They are meant to be jokes.

The first is obviously replacing the c^2 in einstein's equation with the a^2+b^2 it equals according to the pythagorean theorem.

In the second, the top says sin x, and the bottom is n. Rearrange to get (six n)/n. Limit as n goes to infinity is therefore six.

The first is obviously replacing the c^2 in einstein's equation with the a^2+b^2 it equals according to the pythagorean theorem.

In the second, the top says sin x, and the bottom is n. Rearrange to get (six n)/n. Limit as n goes to infinity is therefore six.

Jared Russell

04-17-2003, 08:37 PM

e ^ (pi * i) + 1 = 0

My Calculus teacher gave the above as proof that God exists...

:eek: :ahh:

My Calculus teacher gave the above as proof that God exists...

:eek: :ahh:

Raven_Writer

09-17-2003, 07:12 PM

http://mathworld.wolfram.com/

I'm not sure if it's got "crazy" formulas, but it does have many I never knew existed (even though I'm only in year 3 of integraded math)

I'm not sure if it's got "crazy" formulas, but it does have many I never knew existed (even though I'm only in year 3 of integraded math)

Pat Bogard

09-17-2003, 08:08 PM

This one is used to show enzyme kinetics, its actually derived through about 20 equations

Variables and Explanation of symbols used:

{} = subscripts

[] = concentration

E = enzyme

S = substrate

ES = enzyme-substrate bonded together

[S] = concentration of substrate

K{M} = (k{-1} + k{2})/k{1}

where k{1}, K{-1}, and k{2} are rate constants for the reaction these are dependant on the substrate and enzyme

V{max}= k{2}*[E]{T}

[E]{T} = [ES]

and the equation

Michaelis - Menten Model:

V{0} = V{max}*([S]/[S] +K{M}

Variables and Explanation of symbols used:

{} = subscripts

[] = concentration

E = enzyme

S = substrate

ES = enzyme-substrate bonded together

[S] = concentration of substrate

K{M} = (k{-1} + k{2})/k{1}

where k{1}, K{-1}, and k{2} are rate constants for the reaction these are dependant on the substrate and enzyme

V{max}= k{2}*[E]{T}

[E]{T} = [ES]

and the equation

Michaelis - Menten Model:

V{0} = V{max}*([S]/[S] +K{M}

Gadget470

09-18-2003, 10:09 PM

Originally posted by Abwehr

e ^ (pi * i) + 1 = 0

My Calculus teacher gave the above as proof that God exists...

:eek: :ahh:

Care to explain how?

e ^ (pi * i) + 1 = 0

My Calculus teacher gave the above as proof that God exists...

:eek: :ahh:

Care to explain how?

patrickrd

09-18-2003, 11:12 PM

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

-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

09-19-2003, 05:36 PM

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

George1902

09-19-2003, 08:47 PM

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

(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

(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

12-30-2003, 05:39 PM

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.

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.

01-02-2004, 11:20 AM

-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.

This reminds me of an equation that my friend showed me that related to trebuceuts. The thing was a page long differential equation.

swando

01-07-2004, 02:43 PM

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..

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

01-07-2004, 02:50 PM

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...

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

01-21-2004, 04:23 PM

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

swando

01-21-2004, 04:45 PM

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 :)

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

01-22-2004, 03:51 PM

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.

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);

}

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

04-08-2004, 10:47 PM

Try solving problem #2 in this thread:

http://www.chiefdelphi.com/forums/showthread.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.

http://www.chiefdelphi.com/forums/showthread.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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