Author
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Topic: Math Help - Music of the Spheres
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crashfrog
Member (Idle past 1489 days) Posts: 19762 From: Silver Spring, MD Joined: 03-20-2003
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Message 1 of 10 (247973)
10-01-2005 3:54 PM
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Quick question for the math folks - how do I find the dimensions of a given section of a sphere? If I have a sphere of radius s, and it intersects a plane at distance d from the center of the sphere, what is the radius of the circle formed at the intersection? Sort of like this: A highly practical question for me this weekend.
Replies to this message: | | Message 2 by Chiroptera, posted 10-01-2005 4:13 PM | | crashfrog has not replied | | Message 3 by nwr, posted 10-01-2005 4:15 PM | | crashfrog has not replied | | Message 4 by Ben!, posted 10-01-2005 4:20 PM | | crashfrog has not replied | | Message 7 by RAZD, posted 10-01-2005 8:27 PM | | crashfrog has not replied |
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Chiroptera
Inactive Member
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Message 2 of 10 (247977)
10-01-2005 4:13 PM
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Reply to: Message 1 by crashfrog 10-01-2005 3:54 PM
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quote: If I have a sphere of radius s, and it intersects a plane at distance d from the center of the sphere, what is the radius of the circle formed at the intersection?
r = \sqrt{s^2 - d^2}. I think the area of the "cap" formed is A = 2 \pi (s^2 - sd) but someone better check my work here. Note: In case you can't read TeX, \sqrt is the square root sign, with the argument in the curly brackets. \pi is 3.14159....
This message is a reply to: | | Message 1 by crashfrog, posted 10-01-2005 3:54 PM | | crashfrog has not replied |
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nwr
Member Posts: 6409 From: Geneva, Illinois Joined: 08-08-2005 Member Rating: 5.3
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Message 3 of 10 (247979)
10-01-2005 4:15 PM
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Reply to: Message 1 by crashfrog 10-01-2005 3:54 PM
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You have a line of length d from the center of the sphere to the center of the intersecting circle. A line of length r, where r is the radius of the intersecting circle, joins the center of that circle to a point on its circumference. Those two lines are perpendicular, so are the legs of a right angled triangle. The hypotenuse is a line connecting the center of the sphere to the point on the circumference. That gives a length s for the hypotenuse. The Pythagorian theorem gives us d^2 + r^2 = s^2. I hope that helps. edit: corrected typo. This message has been edited by nwr, 10-01-2005 03:16 PM
This message is a reply to: | | Message 1 by crashfrog, posted 10-01-2005 3:54 PM | | crashfrog has not replied |
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Ben!
Member (Idle past 1420 days) Posts: 1161 From: Hayward, CA Joined: 10-14-2004
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Message 4 of 10 (247981)
10-01-2005 4:20 PM
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Reply to: Message 1 by crashfrog 10-01-2005 3:54 PM
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quickie
crash, ... Did this off the cuff. I'm no mathematician. Here's what I did: 1. Since spheres are symmetric in 3D space, I can validly rotate my sphere / plane system to my coordinate system however I want, and not change the problem. 2. I chose to orient the sphere / plane system so that the center of my sphere is at the CENTER of my XYZ coordinate axes, and so that the plane is perpendicular to XY plane AND perpendicular to the X axis (parallel to the Y axis). Like I said, I can do this because of symmetry. This makes the problem turn into a completely 2D problem: 3. This made things SUP easy; easy geometry / trig: circle function: x2 + y2 = r2 so for point p, px2 + py2 = r2 or d2 + r_12 = r2 so
r_1 = sqrt( r2 - d2 ). At least, that's what I came up with. Ben AbE: crap I did a lot of work that the "smart" people didn't have to do oh well. This message has been edited by Ben, Saturday, 2005/10/01 01:21 PM This message has been edited by Ben, Saturday, 2005/10/01 01:23 PM
This message is a reply to: | | Message 1 by crashfrog, posted 10-01-2005 3:54 PM | | crashfrog has not replied |
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ringo
Member (Idle past 434 days) Posts: 20940 From: frozen wasteland Joined: 03-23-2005
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Message 5 of 10 (247984)
10-01-2005 4:51 PM
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Reply to: Message 4 by Ben! 10-01-2005 4:20 PM
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Re: quickie
Ben writes: I did a lot of work that the "smart" people didn't have to do Don't feel bad - that's the way we used to do it (on the cave walls) back in high school.
People who think they have all the answers usually don't understand the questions.
This message is a reply to: | | Message 4 by Ben!, posted 10-01-2005 4:20 PM | | Ben! has not replied |
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Chiroptera
Inactive Member
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Message 6 of 10 (247988)
10-01-2005 5:01 PM
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Reply to: Message 4 by Ben! 10-01-2005 4:20 PM
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Re: quickie
quote: crap I did a lot of work that the "smart" people didn't have to do
I don't know whether I'm one of the "smart people", but your method is exactly the one I used. I just didn't post (or describe) my sketch. Edited to make my meaning more clear. This message has been edited by Chiroptera, 01-Oct-2005 09:12 PM
This message is a reply to: | | Message 4 by Ben!, posted 10-01-2005 4:20 PM | | Ben! has not replied |
Replies to this message: | | Message 8 by crashfrog, posted 10-01-2005 10:51 PM | | Chiroptera has not replied |
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RAZD
Member (Idle past 1427 days) Posts: 20714 From: the other end of the sidewalk Joined: 03-14-2004
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Message 7 of 10 (248041)
10-01-2005 8:27 PM
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Reply to: Message 1 by crashfrog 10-01-2005 3:54 PM
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now that you have the answer
can you tell us why?
This message is a reply to: | | Message 1 by crashfrog, posted 10-01-2005 3:54 PM | | crashfrog has not replied |
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crashfrog
Member (Idle past 1489 days) Posts: 19762 From: Silver Spring, MD Joined: 03-20-2003
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Message 8 of 10 (248077)
10-01-2005 10:51 PM
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Reply to: Message 6 by Chiroptera 10-01-2005 5:01 PM
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Re: quickie
Damn, yeah, that's the way I was trying to do it in my head, but I just couldn't fill in the algrebra for some reason. (Well, not "some reason" - the reason is that I suck at math, hardcore.) As to what I'm doing - anybody ever heard of "casemodding"? I'm cutting a hole in my computer so that a plasma globe I bought will project out of the front of the case, and I want it to be nice and snug. The result should be something akin to this: Which is where I got the idea. Anyway, thanks to everybody that replied. Now that you've laid it out for me, I'm embarassed that its so simple.
This message is a reply to: | | Message 6 by Chiroptera, posted 10-01-2005 5:01 PM | | Chiroptera has not replied |
Replies to this message: | | Message 9 by gene90, posted 10-02-2005 2:48 PM | | crashfrog has replied |
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gene90
Member (Idle past 3845 days) Posts: 1610 Joined: 12-25-2000
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Message 9 of 10 (248226)
10-02-2005 2:48 PM
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Reply to: Message 8 by crashfrog 10-01-2005 10:51 PM
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Re: quickie
Cool. Are you going to have to shield it? My understanding is that plasma globes are powered by small Tesla coils.
This message is a reply to: | | Message 8 by crashfrog, posted 10-01-2005 10:51 PM | | crashfrog has replied |
Replies to this message: | | Message 10 by crashfrog, posted 10-02-2005 8:47 PM | | gene90 has not replied |
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crashfrog
Member (Idle past 1489 days) Posts: 19762 From: Silver Spring, MD Joined: 03-20-2003
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Message 10 of 10 (248351)
10-02-2005 8:47 PM
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Reply to: Message 9 by gene90 10-02-2005 2:48 PM
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Re: quickie
Are you going to have to shield it? I doubt it. I don't think that guy did. I'm no eletrical engineer, but I suspect several things are working in my favor: 1) My case is a big grounded cage, basically. 2) It's big enough that the inverter circuit for the globe will be fairly far from any crucial components. 3) The inverter/transformer assembly is completely encased in this black plastic stuff. I think I'll be ok. I doubt this thing kicks out any worse RF than neon tubes, or florescent lights, or those cold-cathode kits that folks throw right into their cases. Should look pretty sweet. I'm keeping a build log; interested folks can catch it on my (rarely-updated) blog You Magnificent Bastard.
This message is a reply to: | | Message 9 by gene90, posted 10-02-2005 2:48 PM | | gene90 has not replied |
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