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**otg****Member**- Registered: 2013-03-25
- Posts: 7

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.

*Last edited by otg (2013-03-25 16:02:54)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,972

hi otg

Welcome to the forum.

Have a look at the picture below. Is it correct?

I think I can generate the right equations. The red curve is meant to show where the two solids intersect. At those points both the equation for the surface points of a cylinder and those for the sphere must be obeyed. I think that is enough to find the diameter of the cylinder. Haven't tried yet as I wanted to be sure I had the picture first.

ps. Just re-read the problem. You want the cylinder to be smaller than the sphere. I'll have another go and post later.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**otg****Member**- Registered: 2013-03-25
- Posts: 7

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,972

hi otg

Thanks for the pictures. I've taken the first and zoomed in on the intersection.

There are 4 significant points where the sphere and the cylinder touch, ADBC and they form the ellipse. You cannot see D because it is exactly hidden behind C

Picture 1 below shows the zoom with labels.

Picture 2 shows the ellipse.

If that all looks OK, post back and I'll start making the equations.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**otg****Member**- Registered: 2013-03-25
- Posts: 7

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.)Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,972

hi otg

Flat surface or two spheres makes a circle? YES.

cylinder / sphere makes an ellipse. YES. Maybe a circle if the radii are carefully chosen.

The diagram below (not to scale) shows a front view and a plan view of the solids. I've imagined the cylinder turned until its axis is vertical and for the front view imagine we have then sliced through the sphere at its centre. We are seeing the true length of AB = 1.75 The midpoint of AB I have called E, so AE = 1.75/2 = 0.875

O is the centre of the sphere so we know OA = 6.5/2 = 3.25

So I can use Pythagoras theorem to calculate OE.

The plan view makes the cylinder look like a circle, because we are looking straight down on it. P is any point on the axis of the cylinder, so PE on the front view will give us the radius of the cylinder.

On the plan view you can see where C and D actually are. So I have put a dotted line straight down to show where C is on the front view. But be careful here. C is not in the plane of the front view, so that distance CP on the front is not the true length. You have to be looking straight down on CD to see its true length, CD = 0.5

The midpoint of CD is F, so CF = 0.5/2 = 0.25 I'm planning to use Pythag again in the triangle CFP so I can calculate the true length of CP.

I think I'll have to use a letter, say r, to represent the radius of the cylinder, ie. CP = r

Also it looks like I'll need two more uses of pythag on triangles OFC and AEP.

Oh boy, this is getting complicated. I think I need to cool my brain for a bit ... got washing up to do ..... so I'll stop there for now and start putting down some numbers / algebra in the next post.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**otg****Member**- Registered: 2013-03-25
- Posts: 7

Yes!

I'm following you so far.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,972

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

Pythag on OFC

Pythag on FPC

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**otg****Member**- Registered: 2013-03-25
- Posts: 7

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...............

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**otg****Member**- Registered: 2013-03-25
- Posts: 7

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.Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,972

hi otg,

You seem to have understood most of it, so that's impressive. The bit you don't understand is due to a bit of algebra.

I'll start with r - a and then put in a value for a.

I've made a diagram below that should help.

I made a square, r by r and took off a smaller amount 'a' from each side. Then I split the square into four parts.

First part is a square r - a on each side, so its area is

That's the area we need.

The second and third parts are rectangles, one is r by a, the other is a by r.

So together they have area

But if you put these three back together, to make the r by r square the rectangles overlap, so I'm counting some area twice.

Can you see that the amount I've got twice is

So to make the original square, r by r, you need

Re-arrange this

It is quite a common mistake to miss out that 2ra term.

Let's try with numbers to make sure we've got the right thing.

Can you see you don't get the right answer if you miss out the 2ra term?

So back to your question.

What is

So here a = 0.11037

I hope that is clear.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**otg****Member**- Registered: 2013-03-25
- Posts: 7

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.

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