math problem: help me figure this out. (now with paypal reward)

[notfred]

Here's the semi functional script

NOTE: This works in Firefox 1.5+ ONLY!

Try setting "number of teeth on cog" to something small (maybe 10) and "number of teeth on chainring" to something large (try 150), and you'll see the appropriate lines drawn in the diagram below.

 

[chuckywang]

Originally posted by: notfred

Originally posted by: chuckywang

Originally posted by: MathMan
Ok, it can be simplified a bit more...


sin (theta + pi/2) = cos (theta), and
cos (theta + pi/2) = sin (theta), so

U = x * sin(theta), x * cos(theta)
V = d + y * sin(theta), y * cos(theta)

But sin(theta) = (y-x)/d, and cos(theta) = sqrt(1 - (y-x)^2 / d^2)), so we have

U = x * (y-x)/d , x * sqrt(1 - (y-x)^2 / d^2))
V = d + y * (y-x)/d, y * sqrt(1 - (y-x)^2 / 2^2))

cos(theta + pi/2) = -sin(theta)

BTW, your solution works. If you want $5, PM me a paypal address I'll post a link to the semi-functional script in a few

Sweet, thanks!

 

[Martin]

Originally posted by: notfred

Originally posted by: chuckywang

Originally posted by: MathMan
Ok, it can be simplified a bit more...


sin (theta + pi/2) = cos (theta), and
cos (theta + pi/2) = sin (theta), so

U = x * sin(theta), x * cos(theta)
V = d + y * sin(theta), y * cos(theta)

But sin(theta) = (y-x)/d, and cos(theta) = sqrt(1 - (y-x)^2 / d^2)), so we have

U = x * (y-x)/d , x * sqrt(1 - (y-x)^2 / d^2))
V = d + y * (y-x)/d, y * sqrt(1 - (y-x)^2 / 2^2))

cos(theta + pi/2) = -sin(theta)

BTW, your solution works. If you want $5, PM me a paypal address I'll post a link to the semi-functional script in a few

I'd be interested in seeing it. I've been doing increasing amounts of javascript at work, and I must say can be quite interesting...

 

[Fenixgoon]

nm, misread question

 

[bluewall21]

My way is cooler.

Assuming I didn't do anything stupid, which is always possible.

 

[notfred]

Originally posted by: Martin

Originally posted by: notfred

Originally posted by: chuckywang

Originally posted by: MathMan
Ok, it can be simplified a bit more...


sin (theta + pi/2) = cos (theta), and
cos (theta + pi/2) = sin (theta), so

U = x * sin(theta), x * cos(theta)
V = d + y * sin(theta), y * cos(theta)

But sin(theta) = (y-x)/d, and cos(theta) = sqrt(1 - (y-x)^2 / d^2)), so we have

U = x * (y-x)/d , x * sqrt(1 - (y-x)^2 / d^2))
V = d + y * (y-x)/d, y * sqrt(1 - (y-x)^2 / 2^2))

cos(theta + pi/2) = -sin(theta)

BTW, your solution works. If you want $5, PM me a paypal address I'll post a link to the semi-functional script in a few

I'd be interested in seeing it. I've been doing increasing amounts of javascript at work, and I must say can be quite interesting...

Two posts above yours

 

[Koing]

ATOT comes up trumps again :thumbsup:

Koing

 

[Pepsei]

wow, the differences between this and the other guy that Loke turned in is night and day.

i guess it's the way you have to present your request, and the fact that we always enjoy notfred's writing in the past.

 

[Koing]

Originally posted by: Pepsei
wow, the differences between this and the other guy that Loke turned in is night and day.

i guess it's the way you have to present your request, and the fact that we always enjoy notfred's writing in the past.

What happened with the other guy?

Koing

 

[notfred]

Originally posted by: Pepsei
wow, the differences between this and the other guy that Loke turned in is night and day.

i guess it's the way you have to present your request, and the fact that we always enjoy notfred's writing in the past.

Yeah, he was trying to get other people to do his homework, and I wasn't.

 

[linkgoron]

err maybe I'm an idiot but cant you you two simple equations, without sin and anything and get it done?
A=radius (left circle)
B=radius (right circle)
using y=mx (that line that connects both circles)
left circle being x^2+y^2=A^2
right circle (x-d)^2+y^2=B^2
and that line Y-MX=0

EDIT 1:
it's kind of complicated, but it's a 5 line answer, I think at least
you use distance between two err dots? (don't know terms in English lol) (0,0) and the left circle's resting(?) (using the sqrt((x1-x2)^2+sqrt(y1-y2))=Distance) and another equastion using the right circle and using this:
|D*M|/SQRT(2) = B

EDIT2:
Well It works, or should work, and I get what you wanted, but it seems you found an answer so NM...

 

[Pepsei]

Originally posted by: Koing

Originally posted by: Pepsei
wow, the differences between this and the other guy that Loke turned in is night and day.

i guess it's the way you have to present your request, and the fact that we always enjoy notfred's writing in the past.

What happened with the other guy?

Koing

i think loke found out where he goes to school and send the link of the thread to the school dean.

 

[The Batt?sai]

better offer more than $5

 

[Theiananator]

6743.73