Need help on Calc (corrected wording)

[Xylitol]

Diagram:
http://img90.imageshack.us/img...025/diagramhelpko4.png

How does the depth relate to the 2R which is a side to the horizontal cross section involving 2R and H?
I don't know if it does relate, but if it does, that'd help a lot

Thanks

 

[blinky8225]

It's R?

 

[Xylitol]

oh shit i drew it wrong
one sec let me put up a new pic

 

[Xylitol]

bump with new picture and right wording

 

[Ronstang]

It makes less sense now.

 

[NoShangriLa]

?

 

[blinky8225]

There's no 2R in your drawing, buddy. And if 2R is width of rectangle, the depth is still R.

 

[Xylitol]

my badddddddd
i didn't edit the wording i guess - only the picture
NOW it should look better

 

[blinky8225]

As far as I know, no relation unless you were given other constraints, arc length, or angle measure.

 

[Xylitol]

anyone else?

 

[frostedflakes]

If you make a right triangle out of d and r, you can approximate the third side of the triangle as the arc length of 1/4 of the circle, which is (pi*r)/2. Using the Pythagorean theorem, you can them find a relationship between side d and the two other sides in terms of r.

Of course (assuming this is right), it's just an approximation, that may not be what your instructor's looking for.

Here's a picture, I know for me it's much more helpful to see something than have somebody try to explain it (screwed up a few things and fixed picture).

http://xs127.xs.to/xs127/08225/calc233232.jpg

 

[QED]

I'm assuming as the depth varies, so does the width of the rectangular cross section.

When the depth is 0, the width of the rectangle is 0.
When the depth is R (i.e. is exactly equal to the radius of the circle), then the width of the rectangle is 2R (since the width of the rectangle is essentially the diameter of the circle).
When the depth is 2R (i.e, at the top of the circle), the width of the rectangle is again 0.

You can actually calculate what the width of the rectangle is for any given depth D using the Pythagoream Theorem. By my calculations, I have the width is equal to 2 times the square root of ( 2DR - D^2 ).

 

[hypn0tik]

To make the problem easier, place your system on Cartesian coordinates. You only need to concern yourself with 2D, as the length of the cylinder doesn't matter.

The cross section of your cylinder is a circle, so you can draw a circle with the equation x^2 + (y-R)^2 = R^2.

When you draw your rectangle, it's a line parallel to the x-axis and will have the equation y = d, where d is the depth of the line. You can easily compute the x-coordinates of the points of intersection of the line and the circle to find the length of the side of your rectangle.

 

[Xylitol]

Originally posted by: QED
I'm assuming as the depth varies, so does the width of the rectangular cross section.

When the depth is 0, the width of the rectangle is 0.
When the depth is R (i.e. is exactly equal to the radius of the circle), then the width of the rectangle is 2R (since the width of the rectangle is essentially the diameter of the circle).
When the depth is 2R (i.e, at the top of the circle), the width of the rectangle is again 0.

You can actually calculate what the width of the rectangle is for any given depth D using the Pythagoream Theorem. By my calculations, I have the width is equal to 2 times the square root of ( 2DR - D^2 ).

ok
I used that to finish up the last 2 questions on my project
Thanks a lot

 

[frostedflakes]

Originally posted by: hypn0tik
To make the problem easier, place your system on Cartesian coordinates. You only need to concern yourself with 2D, as the length of the cylinder doesn't matter.

The cross section of your cylinder is a circle, so you can draw a circle with the equation x^2 + (y-R)^2 = R^2.

When you draw your rectangle, it's a line parallel to the x-axis and will have the equation y = d, where d is the depth of the line. You can easily compute the x-coordinates of the points of intersection of the line and the circle to find the length of the side of your rectangle.

And I think this is the correct answer, so disregard my approximation OP.

 

[Xylitol]

Originally posted by: Xylitol

Originally posted by: QED
I'm assuming as the depth varies, so does the width of the rectangular cross section.

When the depth is 0, the width of the rectangle is 0.
When the depth is R (i.e. is exactly equal to the radius of the circle), then the width of the rectangle is 2R (since the width of the rectangle is essentially the diameter of the circle).
When the depth is 2R (i.e, at the top of the circle), the width of the rectangle is again 0.

You can actually calculate what the width of the rectangle is for any given depth D using the Pythagoream Theorem. By my calculations, I have the width is equal to 2 times the square root of ( 2DR - D^2 ).

ok
I used that to finish up the last 2 questions on my project
Thanks a lot

I don't completely get how you're using pythag to get that
can you or someone else explain in more depth?

I can see the D^2 but I don't get where you got the 2DR from

Thanks

 

[Xylitol]

bump?

 

[Xylitol]

/

 

[QED]

Sorry, I was booted into Vista at the time and didn't have my copy of Photoshop available to draw a picture of how I derived that formula.

I'll try to explain it in text...

Consider just the end circle with radius R. There is a horizontal chord which is distance D from the bottom of the circle. We wish to find
the length of this chord.

Now consider these three points: the center of the circle, and the two endpoints of the chord (which lie on the circle as well). They form an isoscles triangle, with the base of the triange being the chord, and the two other sides being equal to the radius R of the circle.

Now draw a vertical line through the center of the circle. It will bisect the chord, and hence bisect the isocsles triangle into two congruent right triangles.

The base of each right triangle is exactly half of the chord length. The height of the right triangles is equal to the radius R minus the depth D (if D < R), or the depth D minus the radius R (if D>R). The hyptonenuse of each right triangle is simply the radius of the circle R.

Hence, if C is the chord length, then by the Pythagorean Theorem we must have ( C / 2 ) ^ 2 + (D - R) ^2 = R^2. Solve this equation for C, and you'll have your chord length.