NationStates Jolt Archive

Take a cylinder of known dimensions, and extend it by adding circular curves of known radii along there right-angled outer edge circumfrences, thus making a "cheese-wheel" shape, with flat top and bottom and circular side and with curved upper and lower edges, of known dimensions. How do you find the formula or graph of formulae for an ellipsoidal shape of known volume that meets but does not intersect the original shape at tangential circles at its top and bottom circumfrences?

Any help at all will be greatly appreciated. :) Clerification available on request.

WTF?!?!?

Geometry is teh sukc

Take a cylinder of known dimensions, and extend it by adding circular curves of known radii along there right-angled outer edge circumfrences, thus making a "cheese-wheel" shape, with flat top and bottom and circular side and with curved upper and lower edges, of known dimensions. How do you find the formula or graph of formulae for an ellipsoidal shape of known volume that meets but does not intersect the original shape at tangential circles at its top and bottom circumfrences?

Any help at all will be greatly appreciated. :) Clerification available on request.
Aight

WHAT?!?

Cut the cheese?

Sorry--had to get that in before anyone else did. :D

WTF?!?!?

Geometry is teh sukc

YOU are the suck!!!

*runs away weeping bitterly*

Geometry is the coward-field of math.

Throw some calc at me

Take a cylinder of known dimensions, and extend it by adding circular curves of known radii along there right-angled outer edge circumfrences, thus making a "cheese-wheel" shape, with flat top and bottom and circular side and with curved upper and lower edges, of known dimensions. How do you find the formula or graph of formulae for an ellipsoidal shape of known volume that meets but does not intersect the original shape at tangential circles at its top and bottom circumfrences?

Any help at all will be greatly appreciated. :) Clerification available on request.
What is a "cheese-wheel" shape?

What are "circular curves"?

Can you draw this shape on paint or something and upload it?

Aight

WHAT?!?

Actually it would be better described as a tire-car shape, as many cheese-wheels are totally ellipsoidal.

Imagine a tire-like shape of known dimensions with quarter-circular edges curving from its road-surface to its face. Imagine that we wish to fit it inside of an ellipsoid, also of known volume, so that the ring of each of the curved circumfrences meets the inner edge of the ellipsoid tangentially, at two circular lines of circumfrence.

Bolded are the bits I do understand.

Imagine a tire-like shape of known dimensions with quarter-circular edges curving from its road-surface to its face. Imagine that we wish to fit it inside of an ellipsoid, also of known volume, so that the ring of each of the curved circumfrences meets the inner edge of the ellipsoid tangentially, at two circular lines of circumfrence.

Too hard for me when I'm drunk

Geometry is the coward-field of math.

Throw some calc at me
.....Well then answer his question Hero.

What is a "cheese-wheel" shape?

What are "circular curves"?

Can you draw this shape on paint or something and upload it?

I can but not readily. I suck at Paint.

OK imagine a tire-shape instead. Or a cheese-wheel with flat sides. Draw a line from the center of one circular side to that of the other. The line describes a "U" shape, with a flat bottom for the "U". The circular edges are the curves of the "U", continued all along the circumfrence.

Or imagine a cylinder, and place a sphere with its center point on a point of one circumfrence of the cylinder, and now place similar spheres on every (one of the infinite) points along that circumfrence of the cylinder, so that it makes a circular ring with a curved edge. Now do this to the other circumfrence of the cylinder, and place flat surfaces between the two rings, on the top and bottom, and a flat ring between them along the circular side of the cylinder.

So now you have a tire-like shape, or a cheese-wheel with a flattened side.

Dru, I am picturing A torus shape, on the end of a cylinder I am afraid I am not "seeing" what you are saying.

A CILINDER WITH ROUNDED EDGES!

Jeez

Geometry is the coward-field of math.

Throw some calc at me

New nick for KEHLPTOPIA? (sp?) ;)

This is not a test for you, this is something I need help with. If you find it so easy, I thankfully anticipate your aid.

A CILINDER WITH ROUNDED EDGES!

Jeez

Well yes but as a cylinder has rounded sides already, I thought a more detailed description was in order. Sorry.

Dru, I am picturing A torus shape, on the end of a cylinder I am afraid I am not "seeing" what you are saying.

Sounds good, except it's a closed torus... not like a standard donut but like a filled one. And on both top and bottom of the cylinder. And the outer edge of the ring is indeed flat, still a cylinder except for the curves on the top and bottom.

Sorry if my descriptive powers are overly complex &/or insufficient.

New nick for KEHLPTOPIA? (sp?) ;)

This is not a test for you, this is something I need help with. If you find it so easy, I thankfully anticipate your aid.

Sorry, Geometry is pretty much useless to my major, and most others.

Bolded are the bits I do understand.

Hopefully by now I have clerified what the internal shape looks like. Now, an ellipsoid ... shaped like an Ibuprofin or a flying saucer, or maybe (probably not) like a rugby ball. We fit the tire-shape inside of that, so that its curved upper and lower circumfrences meet the inner shape tangentially, in two rings, entirely containing the tire-shape inside of the ellipsoid.

Given a known volume for the ellipsoid, and known dimensions of the wheel, how do we calculate the axes of the ellipsoid?

Umm, shouldn't you be doing your homework without the help of the online community?

It's not homework.

To make sure I understand: You're describing the surface of revolution of a "rectangle" with rounded corners? (The rounded corners being quarter circles). That is, if you take a cross section through the center of the thing, you get a rounded "rectangle".

To make sure I understand: You're describing the surface of revolution of a "rectangle" with rounded corners? (The rounded corners being quarter circles). That is, if you take a cross section through the center of the thing, you get a rounded "rectangle".

Yes.

To make sure I understand...

What's wrong with "pretend it's an ellipse and integrate from one flat edge to the other, make 3D?"
That makes it one-variable integration, and if you know what a Torus is then you should know basic conics...

To make sure I understand...

What's wrong with "pretend it's an ellipse and integrate from one flat edge to the other, make 3D?"
That makes it one-variable integration, and if you know what a Torus is then you should know basic conics...

Edit- Err, removing something when I should have added something...
Still the same basic idea- Integrate a sphere/ellipsoid, add a cylinder and think of it as "in between the halves of the sphere (caps/ends)."

Okay. So, you know the dimensions, radius and height, of the original cylinder. You also know the radii of the circular curves that make up the curved edges of the new "cheese-wheel," right?

So, if Rcyl is the radius of the cylinder, Rcir is the radius of the circles (forgive me, I have no subscript), and h is the height of the cylinder, it is possible to find the slope at any point along those curved edges.

For the sake of convenience, we should position our cheese-wheel with its circular center and vertical midpoint at the origin. Then one of our mini-circles (on the x-y plane cross-section of the cheese-wheel) is centered at (Rcyl, h/2 - Rcir). So the equation of that curve is (x - Rcyl)^2 + [y - (h/2 - Rcir)]^2 = (Rcir)^2 (defined only when x > Rcyl, y > (h/2 - Rcir)). Note that this is the upper-right hand edge of the cross-section.

Is that enough of a start for you?

Mycon: Looks like it's my turn to be confused. :confused:

Anyway, yeah... take a rectangle with circular corners and fit it inside of an ellipse so that it meets tangentially at four points, one in each quadrent. Now an infinite number of ellipses will fit this way, but provided that the rotation of the ellipse has a known volume, there will be at most two such ellipses, possibly one, possibly none if the ellipse is not large enough.

Okay. So, you know the dimensions, radius and height, of the original cylinder. You also know the radii of the circular curves that make up the curved edges of the new "cheese-wheel," right?

So, if Rcyl is the radius of the cylinder, Rcir is the radius of the circles (forgive me, I have no subscript), and h is the height of the cylinder, it is possible to find the slope at any point along those curved edges.

For the sake of convenience, we should position our cheese-wheel with its circular center and vertical midpoint at the origin. Then one of our mini-circles (on the x-y plane cross-section of the cheese-wheel) is centered at (Rcyl, h/2 - Rcir). So the equation of that curve is (x - Rcyl)^2 + [y - (h/2 - Rcir)]^2 = (Rcir)^2 (defined only when x > Rcyl, y > (h/2 - Rcir)). Note that this is the upper-right hand edge of the cross-section.

Is that enough of a start for you?

Yes. But... the mini-circle is centered at simply (R(cy), h/2), and the equation of the curve is (x - r(cy))^2 + (y - h/2) = (R(ci))^2 .

Which I have long-since gotten. It's figuring the axes of an ellipse which, when rotated, has a known volume, and which interects the curve(s) tangentially that I am having problems with.

Yes. But... the mini-circle is centered at simply (R(cy), h/2), and the equation of the curve is (x - r(cy))^2 + (y - h/2) = (R(ci))^2 .

Which I have long-since gotten. It's figuring the axes of an ellipse which, when rotated, has a known volume, and which interects the curve(s) tangentially that I am having problems with.

NOPE. The curve starts at (Rcyl, h/2) -- it has to connect with the cylinder -- so its center has to be one radius below that.
:)

NOPE. The curve starts at (Rcyl, h/2) -- it has to connect with the cylinder -- so its center has to be one radius below that.
:)


Yes but you said that it was CENTERED there, using h as the hight of the ORIGINAL cylinder. ;)

The curve RUNS from (R(cy), h/2 + R(ci)) to (R(cy) + R(ci), h/2).

Anyway thanks all for trying and continuing to. :) It's abstract enough when you can see it on paper.

Oh to be clear: the highest point on the curved edge is R(ci) above the top of the original cylinder, which is itself at h/2 .

Yes but you said that it was CENTERED there, using h as the hight of the ORIGINAL cylinder. ;)

The curve RUNS from (R(cy), h/2 + R(ci)) to (R(cy) + R(ci), h/2).

Ah! I misunderstood the picture, then. I thought you meant that the curves began at the upper lower circumferences.

So if what you're saying now is right, the height of the "cheese-wheel" is h+2Rcir?

Edit: Got it, from your last post. That does simplify things.

Ah! I misunderstood the picture, then. I thought you meant that the curves began at the upper lower circumferences.

So if what you're saying now is right, the height of the "cheese-wheel" is h+2Rcir?

Indeed. :) And its radius is R(cy) + R(ci) .

Okies.
Drop down a dimension and look at a cross-sectional view right through the center. (the rounded rectange). Put the origin at the center for convenience, and have the y-axis be the center of rotation for the tire-shape.

Let (X,Y) be the center of the arc in the first quadrant and R be its radius. Thus the the equation for the circle the arc is part of is:
(x-X)² + (y-Y)² = R²

and expressed as a function, the equation for the arc is:
y = Y + √(R²- (x-X)²)

Let's call this function f(x), with first derivative f'(x)

The equation for an ellipse centered on the origin is:
x²/A + y²/B = 1

the top half expressed as a function is:
y = B√(1-x²/A)

Let's call this function g(x), with first derivative g'(x)

The volume of the resulting ellipsoid is (4π/3)A√B
Let the point of intersection be (v,w)
Then, the constraints of the problem are:
f(v) = g(v) (the ellipsoid touches the wheel-shape)
f'(v) = g'(v) (the two are tangential)
V = (4π/3)A√B (volume is as given)

So what you want to do is solve this system of equations for A, B in terms of X, Y, R, and V. Good luck; it's going to be really messy.

P.S.
sorry for using different notation; posts crossing in the night and all that. The key idea is you can use derivatives to express the tangential constraint. But chances are it's going to be messy any way you do it.

Okies.

. . .

Good luck; it's going to be really messy.

Yup. Hence my quandry.

Anyway I gtg2 work. Thanks folks.

The algebra and derivation messiness is all that's holding you back? Use Maple or Mathematica.

Okay, then...
Image A (http://www.northcountrycybermall.com/heritage/hchwheel.jpg) (Truncated ellipse with noticable eccentricity)

Image B (http://www.math.union.edu/software/StageTools/StageManager/Samples/torus/Torus-1.jpg) (if it were lacking the hole in the middle)? (truncated sphere)

Image C (http://www.pillspy.com/img/pill.gif) (Two halves of a sphere/Ellipse connected by a cylinder)
or
Image D (http://www.pill-price.com/pics/zoloft_100mg.gif) (Rectangular prism, with arches added but not covering the edges)?

Also, who would willingly use Maple? It is the spawn of Satin (echo...). (Anyone else here remember Blitz's Klassic Kartoons? Especially Kat & Mus or Edumacation?)

I could learn Maple or somesuch but I'd rather learn the math.

Anyway thanks Dettibok... but isn't that supposed to be V = (4π/3)A²B? :confused: I may just be confused...

Okay, then...
Image A (http://www.northcountrycybermall.com/heritage/hchwheel.jpg) (Truncated ellipse with noticable eccentricity)

Image B (http://www.math.union.edu/software/StageTools/StageManager/Samples/torus/Torus-1.jpg) (if it were lacking the hole in the middle)? (truncated sphere)

Image C (http://www.pillspy.com/img/pill.gif) (Two halves of a sphere/Ellipse connected by a cylinder)
or
Image D (http://www.pill-price.com/pics/zoloft_100mg.gif) (Rectangular prism, with arches added but not covering the edges)?

Also, who would willingly use Maple? It is the spawn of Satin (echo...). (Anyone else here remember Blitz's Klassic Kartoons? Especially Kat & Mus or Edumacation?)

Like the first, but as a perfect circle, and with the edges, which appear at the scale of your picture to form right angles, actually being circular curves. (I am sure that they actually are roughly circular in your picture but this is only due to the scale.)

Thanks all again. Anyway this is something I have had on the shelf for a while. The other night while some tard was claiming that men are better than women at math and spending 100+ posts refusing to solve a problem that a female poster offered him, I originally posted the first post of this thread, just for kicks. ;) But anyway in the hopes of getting some help I started this one.

Thanks especially Dettibok who has put forth the most. Unfortunately there are some flaws in his/er steps. One:

y = B√(1-x²/A)

No...

y = B√(1-x²/A²)

...since actually ('nother error) x²/A² + y²/B² = 1

Two:

The volume of the resulting ellipsoid is (4π/3)A√B

No (as far as I recall)...

V = (4π/3)A²B

I hope I don't sound too critical, you having stooped to my aid and all. :)



...but anyway putting this out here was a bit of a lark. I had already long since gotten it to this point. Where I am stumped is on the messiness to come. :(

Thanks, all, again. :)

have you built a model?

have you built a model?

No... think it would help?

No... think it would help?

i dont really know anything about geometry, but it might give you a better reference than just numbers and lines.

who knows, maybe you just need a fresh look at it

I'd suggest you hurry over to Wolfram maths site. That's where I go when I'm stuck.
http://mathworld.wolfram.com/topics/SolidGeometry.html

This should help.

I think this is the page you need:
http://mathworld.wolfram.com/SolidofRevolution.html
To find the volume of a solid of rotation by adding up a sequence of thin cylindrical shells...
I can't type in the rest cause it has square root signs and such like.