IAteYourMother
Lifer
- Nov 3, 2004
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Theta is the angle of the line perpenicular to the line of tangency that connects the two circles.
Do you want the solution, or just the answer? I have an answer in my post above. If you can verify it by plugging in some numbers, I would like to know if it is correct. If my answer turns out to be correct, I can tell you the solution as well.
How about instead of asking what theta is then you ask what you really wanted? ****** man do your own god damn work. I was trying to get you going in the right direction, work out your own solution.
I don't actually have any example data to test this with. What I'd end up doing is plotting the results and seeing if they look right. I'm wondering how the heck I'm supposed to implement your solution in Javascript, too
Also, I haven't taken calculus in a long time, and I'm fairly rusty.
Originally posted by: notfred
I don't actually have any example data to test this with. What I'd end up doing is plotting the results and seeing if they look right. I'm wondering how the heck I'm supposed to implement your solution in Javascript, too
Also, I haven't taken calculus in a long time, and I'm fairly rusty.
My solution doesn't use any calculus. Just a bit of trig.
Originally posted by: chuckywang
My solution doesn't use any calculus. Just a bit of trig.
Oops. I saw xt/d and it looked like "dx/dt"
I'll try your solution out.
Ok, took me a few minutes to figure it out, but here goes:
Let's call the point where the line y=0 and the mutually tangent line intersect "Z".
Let's call the origin of the small circle on the left "A"-- it should be (0,0).
Let's call the origin of the large circle on the right "B" -- it should be (d, 0).
Now the coordinate for Z is (-e,0) for some positive value of e.
Now the right triangle ZAU is similiar to triange ZBV. In fact, they share the same angle at Z-- let's call this angle theta.
Hence, the sine of theta is equal to x/e. It is also equal to y/(d+e).
We therefore have x/e = y/(d+e), or xd+xe=ye, or e=xd/(y-x) .
Now that we know e, we can solve for theta since we know sin of theta is equal to x/e.
Theta equals the inverse sine of x/e, or the inverse sine of x/((xd)/(y-x), or the inverse sine of (y-x)/d.
Now that we know theta, we just use simple trig to find the coordinates of U and V:
U = x * cos(theta + pi/2), x * sin(theta + pi/2)
V = d + y * cos(theta + pi/2), y * sin(theta + pi/2)
where theta is the inverse sine of (y-x)/d.
This could probably simplified even further using a few trig identities...
Let's call the point where the line y=0 and the mutually tangent line intersect "Z".
Let's call the origin of the small circle on the left "A"-- it should be (0,0).
Let's call the origin of the large circle on the right "B" -- it should be (d, 0).
Now the coordinate for Z is (-e,0) for some positive value of e.
Now the right triangle ZAU is similiar to triange ZBV. In fact, they share the same angle at Z-- let's call this angle theta.
Hence, the sine of theta is equal to x/e. It is also equal to y/(d+e).
We therefore have x/e = y/(d+e), or xd+xe=ye, or e=xd/(y-x) .
Now that we know e, we can solve for theta since we know sin of theta is equal to x/e.
Theta equals the inverse sine of x/e, or the inverse sine of x/((xd)/(y-x), or the inverse sine of (y-x)/d.
Now that we know theta, we just use simple trig to find the coordinates of U and V:
U = x * cos(theta + pi/2), x * sin(theta + pi/2)
V = d + y * cos(theta + pi/2), y * sin(theta + pi/2)
where theta is the inverse sine of (y-x)/d.
This could probably simplified even further using a few trig identities...
Ok, took me a few minutes to figure it out, but here goes:
Let's call the point where the line y=0 and the mutually tangent line intersect "Z".
Let's call the origin of the small circle on the left "A"-- it should be (0,0).
Let's call the origin of the large circle on the right "B" -- it should be (d, 0).
Now the coordinate for Z is (-e,0) for some positive value of e.
Now the right triangle ZAU is similiar to triange ZBV. In fact, they share the same angle at Z-- let's call this angle theta.
Hence, the sine of theta is equal to x/e. It is also equal to y/(d+e).
We therefore have x/e = y/(d+e), or xd+xe=ye, or e=xd/(y-x) .
Now that we know e, we can solve for theta since we know sin of theta is equal to x/e.
Theta equals the inverse sine of x/e, or the inverse sine of x/((xd)/(y-x), or the inverse sine of (y-x)/d.
Now that we know theta, we just use simple trig to find the coordinates of U and V:
U = x * cos(theta + pi/2), x * sin(theta + pi/2)
V = d + y * cos(theta + pi/2), y * sin(theta + pi/2)
where theta is the inverse sine of (y-x)/d.
This could probably simplified even further using a few trig identities...
Now send me my money
Let's call the point where the line y=0 and the mutually tangent line intersect "Z".
Let's call the origin of the small circle on the left "A"-- it should be (0,0).
Let's call the origin of the large circle on the right "B" -- it should be (d, 0).
Now the coordinate for Z is (-e,0) for some positive value of e.
Now the right triangle ZAU is similiar to triange ZBV. In fact, they share the same angle at Z-- let's call this angle theta.
Hence, the sine of theta is equal to x/e. It is also equal to y/(d+e).
We therefore have x/e = y/(d+e), or xd+xe=ye, or e=xd/(y-x) .
Now that we know e, we can solve for theta since we know sin of theta is equal to x/e.
Theta equals the inverse sine of x/e, or the inverse sine of x/((xd)/(y-x), or the inverse sine of (y-x)/d.
Now that we know theta, we just use simple trig to find the coordinates of U and V:
U = x * cos(theta + pi/2), x * sin(theta + pi/2)
V = d + y * cos(theta + pi/2), y * sin(theta + pi/2)
where theta is the inverse sine of (y-x)/d.
This could probably simplified even further using a few trig identities...
Now send me my money
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))
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))
This problem is really simple if you remove all the extra information given.
You have two similar right triangles, where the right angle is formed by the tangent line x and line uv, and similarly for line y and line uv.
Because the two triangles are similar (prove it to yourself and convince yourself that it is), I'll just talk about the smaller of the two triangles to avoid confusion.
The angle theta is the angle between the tangent line x and the perpendicular line x with respect to the x-axis. From the first diagram of the OP, theta looks to be about 20degrees.
Then sin(90 - theta)*x gives you the relative location of the y-component of U, and cos(90 - theta)*x gives you the relative location of the x-component of U.
The same concept can be applied to the larger circle because the two triangles are similar.
And theta is described as above!
Now that you have their relative location, figuring out their absolute location is simple.
Take the smaller circle to be the origin. Then u = (sin(90-theta)*x , -cos(90-theta)*x)
and v = (sin(90-theta)*y , y -cos(90-theta)*y)
Now where's my $5 dammit. I have to eat tommorow!
You have two similar right triangles, where the right angle is formed by the tangent line x and line uv, and similarly for line y and line uv.
Because the two triangles are similar (prove it to yourself and convince yourself that it is), I'll just talk about the smaller of the two triangles to avoid confusion.
The angle theta is the angle between the tangent line x and the perpendicular line x with respect to the x-axis. From the first diagram of the OP, theta looks to be about 20degrees.
Then sin(90 - theta)*x gives you the relative location of the y-component of U, and cos(90 - theta)*x gives you the relative location of the x-component of U.
The same concept can be applied to the larger circle because the two triangles are similar.
And theta is described as above!
Now that you have their relative location, figuring out their absolute location is simple.
Take the smaller circle to be the origin. Then u = (sin(90-theta)*x , -cos(90-theta)*x)
and v = (sin(90-theta)*y , y -cos(90-theta)*y)
Now where's my $5 dammit. I have to eat tommorow!
Yep, after I did a little simplification I see they are pretty much the same. Which is good, I suppose.
In any case, notfred can at least see the logic used to derive the answer...
Oops.. missed the negative sign there... I should've known that since the x-coordinate of U should be negative and x * (y-x)/d is not negative.So then
U = x * (x-y)/d, x * sqrt(1 - (y-x)^2 / d^2 ))
V = d + y * (x-y)/d, y * sqrt(1 - (y-x)^2/d^2))
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 Originally posted by: notfred
BTW, your solution works. If you want $5, PM me a paypal address
I see....
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