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**Posts by otg**

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Yes, Bob, it's the algebra I have difficulty with. But in addition, I don't see how you can look at what I originally described in my first post and be able to know each of the pieces that have to be solved and the formulas needed. The right triangle I can remember. But I am really very impressed with your knowing how to solve this -- and then being able to explain it in a way that a dunce like me can understand it.

Yes, you made it very clear. I used your demonstration of why the formula worked and drew in red the corresponding squares/rectangles in your original sketch (i47.tinypic.com/9tehq9.jpg) so I can easily see how it applies.

Now for a practical application, I'll try to change the sphere's diameter and see if I can solve it. I'll be back later when I've had a chance to work on it. Right now, I have a few things to do around the house.

Thanks again.

bob bundy wrote:

Oh good. I must have been making some sort of sense then. Brain now re-freshed. I have a plan.

I decided to re-do the diagram. Much the same but I've made the front and plan views line up properly and I've re-named C on the front view as F because then each view only has letters for lengths that we are seeing as true (not at a misleading slant angle)

recap

radius of sphere = OA = 3.25 radius of cylinder = CP = r

AE = half major axis of ellipse = 0.875 FC = half minor axis of ellipse = 0.25

Pythag on OAE

(purple triangle on linked image below)Pythag on OFC

(blue triangle)Pythag on FPC

(green triangle)square both sides

cancel the r squared and re-arrange

I will just check this ........

EDIT: checked. I cannot find any errors. Value seems small but fits with the original diagram where it was small.

You asked for the diameter. I'll leave that as an exercise for you.

Bob

Thanks, Bob, for all the work you put into this. With my math at a deficit, I tried reviewing some of the subjects available here to try and refresh my memory and understand each of the steps you illustrated. I also added the colored triangles in this image (*i46.tinypic.com/demija.jpg*) so I could better visualize what you were doing, and then I added the values of the triangle sides as they were calculated. *(This was just a visual aid for me to try and avoid confusion in going between the two different views.)*

I understand everything up to the point where you squared both sides of

and get . But I still can't figure out where the comes from that you added to that step. I know its because of my deficit in understanding, and I know that the resulting cylinder diameter of 0.67663054... should be correct because on paper I drew to scale the sphere and cylinder (illustrated here: i49.tinypic.com/2mp0761.jpg) and can see that it would be correct. But I would like to be able to understand the final steps so I can try it with other diameter spheres.Thank you so much, Bob. I printed it so I can make sure I understand it all. Not that I doubt it at all, I just want to make sure I do understand it. I'm amazed that it could be solved with Pythagoras. I thought it would be a lot more complicated than that. But it just shows how a brilliant mind can "cut through the weeds" to find only the necessary information. Thanks!

Ya' know... I worked years ago as a helper for a land surveyor. A simple trick he taught me was in finding a 90 degree angle to spot on a straight line we were following (although I doubt he knew anything about Pythagorean Theorem). We would measure forward 4-feet and mark the spot. From that spot we would hold a measure of 5-foot and intersect it with another measure of 3-feet from the original spot. He knew it worked, but he didn't know why.

I am now going to go have dinner while I look over my printout and try to tell my wife that I really can listen to her while I look at something else.

I'll be back later...............

Yes!

I'm following you so far.

That's exactly it! If a flat surface were to intersect with the sphere, or two spheres with one another, the intersection would be a circle. Yes?

But a cylinder would make the intersection an ellipse. Right?

An intersection would be symbolized as:

, I believe. (Thought I'd try using what I've learned here so far about LaTex and see if it works.)Thanks, Bob.

I suppose a picture would have helped. I'll try this and see if it works.

(Oops. Looks like newbies can't post pictures here.) Here are links to two pictures that should help in visualizing the problem:

oi45.tinypic.com/2vug6r5.jpg

oi45.tinypic.com/34nslyt.jpg

**otg**- Replies: 11

Thanks in advance to anyone who might be able to help me. I'm not a mathematician (obviously), or much of an expert at anything worth mentioning. But I've searched all over for a solution, or a way to figure this out, and haven't found it. I ran across this site, and was fascinated with all the information available, but still didn't find an answer. Here's the problem:

I have a sphere with a diameter of 6.5", and it intersects with a cylinder of unknown (but smaller) diameter. However, I know that the resulting elliptical intersection of the sphere and the cylinder is 1.75" x 0.5". Given this information, is it possible to calculate the diameter of the cylinder?

Thanks again.

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