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Registration Date: Apr 2001

Location: San Antonio TX USA

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  Reply with quote "width =" 40 "height =" 22

Today, 02:36

I have a right circular cylinder (circle on the x-z axes) and a plane on the x-z axes. If the plane is tilted 40 degrees from the y-axis to the cylinder, is there a straight and direct (ie simple) solution to find the equation on the curve where the plane intersects the cylinder?

Obviously if the plane was 0 degrees to the axis of the cylinder, the equation would be the equation in the circular form of the circle (x square + y square = constant). But I do not remember from the weak past when I took analytical geometry * how to put the problem together in the first place. In part, it's because I can't remember (nor does Google help me at the moment) how I enter the plane's equation.

Any helpful suggestions, even pointing out a web page that spell my problem out in small words and short sentences, would be appreciated.

* I took "trigonometry and analytical geometry" in the first semester of my senior year of high school. It was in 1976. I haven't used neither trig nor analytical geometry since then …


Glenn —– OTR / L, MOT, Tx













Clinically Insane [19659020] Join Date: June 2001

Location: Chicago, Bang! ! Bang

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  Reply with quote "width =" 40 "height =" 22

Today, at 02:53

If I understand right, this is very fine.

The equation will be for an ellipse.

The minor axis is the diameter of the cylinder.

The main axis is the long side of a right triangle. One side of the triangle is the cylinder diameter. Find the length of the perpendicular side and you are there.








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Registration Date: June 2000

Location: California

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  Reply with quote "width =" 40 "height =" 22

Today, at 02:59

Provided that your aircraft does not cross one of the cylinder ends, your crossing will become an ellipse.

Width of the ellipse will always be the width of the cylinder. Length is all you need. To get the length, look at it from the side (where your plane becomes a line). The cylinder becomes a rectangle, with a line passing through it. Then solve it as a right triangle, where the line is the hypotenuse.

Code:

__________ | | C | / | | / | | / | | / | A | -------- | B | |

A to B = cylinder width.
An angle = 40º
B angle = 90º
C angle = 50º

The rest is trig. If you want to get fancy, you may want to simplify the equation to a general solution to the problem. Ellipse equation = cutting angle and cylinder width plus additional designations. I'm lazy this morning, so you can do the rest.








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Registration Date: Apr 2001

Location: San Antonio TX USA

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  Reply with quote "width =" 40 "height =" 22

Today, 03:03

Thanks! Of course it is an ellipse, but also of course that word just refused to come to the surface of my brain.

And of course you're right. That ellipse is just the thing. There is a right triangle of 40-50-90, which is not one of the "do it in the head" triangles like triangles 45-45-90 and 30-60-90, but it is quite doable.

Thanks for your help. I've always felt that "saying the problem out loud helps you understand it better," but sometimes the right words are hidden by overthinking things. Especially for me.


Glenn —– OTR / L, MOT, Tx











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