OT- Calculate the Hypotenuse of a Triangle Wrapped into a Partial Circle

David_M · Jul 20, 2017

I know there are some good mathematicians on here. What is the length of the hypotenuse of the curved triangle (marked "x")? I can get it by letting the drawing program tell me or by treating it as a partial ellipse, other ways?

edit: It is hard to read the dimensions, they are: Side Adjacent: 7.854 ; Side Opposite: 4.899 ; Hypotenuse of the normal Triangle is: 9.2566 ; The Radius of the curved Triangle is: 5.000 ; and they are Right Triangles.

http://www.pbase.com/image/165865059/original

alextnnr · Jul 20, 2017

is it projected onto a cylinder?

If so you can just use the a^2+b^2=C^2

a=1/4*(2*pi*r)
b= height

This is because you can just treat the lower curved side of the triangle as a 1/4 of the perimeter. it is still a triangle just a curved one.

billmac · Jul 20, 2017

Your drawing is a little ambiguous, but if we assume that you have a helix, then imagine that you took a drawing of your planar right angles triangle and then wrapped it around the circumference of a cylinder of the required radius. Since you haven't stretched anything, the length of the hypotenuse remains the same and you can use plain old Pythagoras as alextnnr suggests.

If you have something more elaborate in mind then we will need to have a bit better definition of your shape.

Rich L · Jul 20, 2017

Yes, your drawing is ambiguous. If the triangle is wrapped (not projected) on a cylinder then use Mr Pythagoras. If you're really asking about the planar triangle (the one with curves) in your drawing then you don't have a right triangle any more. You can make the "long side" (not a hypotenuse any more) what you want and follow any curve you want. I suspect you're wrapped. Yes?

Cheers,
Rich

David_M · Jul 20, 2017

"Yes, your drawing is ambiguous."

To me too. :eek:

This shows it better. It is a right triangle drawn on a cylinder with the hypotenuse formed by removing a planer surface that is angled. I have it tilted forward in this view:
http://www.pbase.com/image/165867863

Showing it with an ellipse formed if the hypotenuse is closed.
http://www.pbase.com/image/165867866

Calculated using 7" x 5" in the sigma summation. the 5" was given and the 7" came from the square root(5"^2 + 4.899"^2). I think the length is 9.4903422325 (a quad of the ellipse).
http://www.pbase.com/image/165867865

WizardOfBoz · Jul 20, 2017

This is exactly like a right triangle. I confirmed that the base is 1/4 of the circumference of a circle of radius 5. Using a^2 + b^2 = c^2, I get 9.2566.
..

EPAIII · Jul 20, 2017

Yes, the drawing has insufficient information. Is it on a cylinder? Or on a sphere?

If it is on a cylinder, it can be un-wound (mentally) and treated like a triangle in a plane.

If it is on a sphere, then it gets a lot more complicated. For one thing, we would need the radius (or diameter) of that sphere.

No sense in knocking ourselves out until the problem is properly stated.

David_M · Jul 20, 2017

I misstated in the title how it is constructed. It is a 10" dia. cylinder (not sphere) with parts trimmed away (using planer surfaces) to leave a triangle.

This may show it best: http://www.pbase.com/image/165869509/large

I unrolled the triangle to show its shape in 2d (the blue outline; red shows an imaginary straight line hypotenuse). You can see that the real hypotenuse is a curved line with a constantly varying radius.

Does this make sense?

billmac · Jul 21, 2017

I think your latest drawing represents a conic section, specifically an ellipse. It would be useful to see a couple of orthogonal views to confirm this. Certainly the shapes produced by a slicing plane through a cylinder or more generally a cone are conic sections, i.e. parabolas, ellipses, hyperbolas.

boslab · Jul 21, 2017

Not as simple as I'd like!
The arc length of a parametrized curve - Math Insight
Mark

billmac · Jul 21, 2017

boslab said:
Not as simple as I'd like!
The arc length of a parametrized curve - Math Insight
Mark

That is for a helix; unfortunately I think the OP has an ellipse, in which case he will need to evaluate an elliptic integral. Probably best to look for a specific calculator to do that rather than first principles.

Rich L · Jul 21, 2017

If it's wrapped on a cylinder (no conic section involved) then it's straight right angle triangle math (Mr P again ). If you have a curved long side ( not a hypotenuse ) to start with in the planar world that is wrapped on a cylinder that's an entirely different problem in which you have to define the length ahead of time. Then when you wrap it on the cylinder that length remains the same.

David_M said:
I misstated in the title how it is constructed. It is a 10" dia. cylinder (not sphere) with parts trimmed away (using planer surfaces) to leave a triangle.

This may show it best: http://www.pbase.com/image/165869509/large

I unrolled the triangle to show its shape in 2d (the blue outline; red shows an imaginary straight line hypotenuse). You can see that the real hypotenuse is a curved line with a constantly varying radius.

Does this make sense?

WizardOfBoz · Jul 21, 2017

If its a right triangle wrapped onto a cylinder, with the base perpendicular to the axis of the cylinder, I think your answer is 9.2566, and the angle is 31 degrees, 57 minutes, and 15 seconds.

David_M · Jul 21, 2017

This shows the steps involved in making the triangle: http://www.pbase.com/image/165872286/large

It ends up being made from one quarter of the circumference of the starting 10" diameter cylinder.

Rich L · Jul 21, 2017

David_M said:
This shows the steps involved in making the triangle: http://www.pbase.com/image/165872286/large

It ends up being made from one quarter of the circumference of the starting 10" diameter cylinder.

If your second planar cut is a perpendicular one then I think your early post with the binomial series calculation will do the trick for the elliptical path as your long side.

EPAIII · Jul 22, 2017

If, as you seem to be stating, it is a right triangle that is just wrapped around the surface of a cylinder, then yes: a^2 + b^2 = c^2

Totally straight forward. And it does not matter what angle it is oriented as you wrap it on that cylinder, it remains the same right triangle that you started with. Try it with a paper triangle and a pipe or some other cylinder.

extropic · Jul 22, 2017

David_M said:
This shows the steps involved in making the triangle: http://www.pbase.com/image/165872286/large

It ends up being made from one quarter of the circumference of the starting 10" diameter cylinder.

This is the information that you should have provided in the original post.

The right triangle is a red herring. It's only significance is that it shares two sides (so, three vertices) with the form that you care about.

The length of "X" is the circumference of the ellipse/4. "X" is not a straight line.

If you don't know how to calculate the circumference of an ellipse, Google it.

David_M · Jul 22, 2017

extropic said:
This is the information that you should have provided in the original post.

The right triangle is a red herring. It's only significance is that it shares two sides (so, three vertices) with the form that you care about.

The length of "X" is the circumference of the ellipse/4. "X" is not a straight line.

If you don't know how to calculate the circumference of an ellipse, Google it.

You are right. It was a learning process for me, I should had thought about it more before I posted. My stating, in the title, that it was wrapped was wrong. :wall:

Rich L · Jul 22, 2017

David_M said:
You are right. It was a learning process for me, I should had thought about it more before I posted. My stating, in the title, that it was wrapped was wrong.

It's OK. We all were trying to wrap our heads around your problem. We just gotta quit calling it a hypotenuse

extropic · Jul 22, 2017

David_M said:
You are right. It was a learning process for me, I should have thought about it more before I posted. My stating, in the title, that it was wrapped was wrong.

It was well worth wading through the trash, in this thread, to read those words.

My sincerest desire is that everyone THINK ABOUT IT before they post.

I know . . . fat chance.

OT- Calculate the Hypotenuse of a Triangle Wrapped into a Partial Circle

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