Challenge(?): Intersection curve between two cylinders in a specific position

[Pier Aiello]

As the figure above shows, we have two identical cylinders, one blue and one black. We want to put the two in the following way: the green line on the black cylinder must do an angle of 45 degrees with the green line on the blue cylinder, and those two lines are touching in one point that is one ending point of the second line. The axes (1) (vertical for the black cylinder, and at 45° degrees for the blue one) are on the same plane, that is, the two lines forming the axes are not skewed, but they are intersected somewhere.

Given that, we have that the blue cylinder is intersecting the black one, and we want to know which is the curve that this intersection produces on the blue cylinder.

How do you solve this with you calculator (2) (? (eventually, are you able to plot the solution and/or the 3d scenario with the curve highlighted?)

Note: (i still have not solved it, i think that it could be a nice challenge)

Some restrictions to avoid tricks:

- The cylinders have an r>0 (in real world can be a tube such as these)

(1) For axis I mean: the line that is normal to the base of the cylinder and that cross the center of that base.
(2) Pick some numbers, like: radius=100 and height=10'000.

Edited: 16 Sept 2013, 6:07 a.m.

[Jean-Michel]

Hello Pier

This is an interesting challenge.
I guess you mean we have to draw the intersection line between the two cylinders, i.e. the curve that is common to both outer surfaces of the cylinders, in plane projection ?
I mean, when the drawing is viewed in a way so that both cylinders appears like rectangles, and not in any projection ? Am I right?

[Pier Aiello]

yes, you can do a plane projection (specifing the plane). Or: Is the curve given by the intersection on a plane (1)? (i don't know actually, even if i guess not)

(1) That is: are all the intersection points on the same plane ?

Edited: 16 Sept 2013, 5:42 p.m.

[peacecalc]

Hello Pier,

that is a funny story. The hobby card modeling has to solve such questions with a lot of intersecting shapes. I wrote for the hp 50g such a program for finding the point whichs belonges to the shapes both. Some of the card modeling collegues thinks that I'm a crazy guy doing this task with on a calculator and not with an 3-D program like rhino or blender.

The main idea of my program is that one surface is build with plane triangles. The other surface is builded with a bunch of lines.

The lines are used to be straight lines which are intersected with the triangles of the other shape. The hp 50g can handle in excellent way vectors and in connection with the dot-product it is easy to find out where the lines and the plane triangles have common points or not.

The program is written in RPL (of course) and a bit long. The documentation is a bit spartan. So I don't know if it make sence to show you here the program text.

Greetings
peacecalc

[Pier Aiello]

Why not? Is just another contribution! Don't feel ashamed.

I'm ashamed here, there are really good people both in using calculators and math! But, imo, it's better to contribute (as best as you can) anyway.

[peacecalc]

Okay Pier,

%%HP: T(3)A(R)F(,);

\<< \-> SPANT1xUMR SPANT2xUMR SPANT1xHULL SPANT2xHULL

  \<< SPANT2xHULL BODY DUP SIZE SWAP SPANT1xHULL SPANT2xHULL - BODY 

	SPANT2xHULL TAIL SPANT2xHULL BODY - 

	SPANT1xHULL BODY SPANT1xHULL TAIL 

	SPANT1xHULL BODY - 

	SPANT2xHULL TAIL SPANT1xHULL BODY - 

	SPANT1xUMR DUP SIZE SWAP SPANT2xUMR SPANT1xUMR - 

	\-> 	HULLxNUM TRAEGxHULL2 RICHT1xHULL2 RICHT2xHULL2 

		TRAEGxHULL1 RICHT1xHULL1 RICHT2xHULL1 

		UMxNUM TRAEGxUM RICHTxUM

    \<< RICHT1xHULL2 RICHT2xHULL2 CROSS RICHT1xHULL1 RICHT2xHULL1 CROSS { } { } 

		\-> NORMxHULL2 NORMxHULL1 SCHNITTxHULL2 SCHNITTxHULL1

      \<< 1 HULLxNUM

        FOR j 1 UMxNUM

          FOR i RICHTxUM i GET NORMxHULL2 j GET DUP ROT DOT DUP

            IF 0, \=/

            THEN SWAP TRAEGxHULL2 j GET TRAEGxUM i GET - DOT SWAP / DUPDUP

              IF 0, \>= SWAP 1, \<= AND

              	THEN    TRAEGxUM i GET 

			RICHTxUM i GET ROT * + DUP 

			TRAEGxHULL2 j GET 

			RICHT1xHULL2 j GET 

			RICHT2xHULL2 j GET 

			NORMxHULL2 j GET 

			\-> SCHNITTxPKT TRAEGxHULL RICHTxHULLx1 RICHTxHULLx2 NORMxHULL

                \<< SCHNITTxPKT TRAEGxHULL - DUP RICHTxHULLx2 CROSS 

					SWAP RICHTxHULLx1 CROSS NORMxHULL DOT 

					SWAP NORMxHULL DOT 

					NORMxHULL NORMxHULL DOT 

					DUP ROT SWAP / UNROT NEG /

                \>> \-> KOEFFxRICHTx1 KOEFFxRICHTx2

                \<< KOEFFxRICHTx1 DUP

                  IF 0, \>= SWAP 1, \<= AND

                  THEN KOEFFxRICHTx2 DUP

                    IF 0, \>= SWAP 1, KOEFFxRICHTx1 - \<= AND

                    THEN j 2 \->LIST SCHNITTxHULL2 SWAP + 'SCHNITTxHULL2' STO

                    ELSE DROP

                    END

                  ELSE DROP

                  END

                \>>

              ELSE DROP

              END

            ELSE DROP DROP

            END

          NEXT

        NEXT 1 HULLxNUM

        FOR j 1 UMxNUM

          FOR i RICHTxUM i GET NORMxHULL1 j GET DUP ROT DOT DUP

            IF 0, \=/

            THEN SWAP TRAEGxHULL1 j GET TRAEGxUM i GET - DOT SWAP / DUPDUP

              IF 0, \>= SWAP 1, \<= AND

              THEN 	TRAEGxUM i GET 

			RICHTxUM i GET ROT * + DUP 

			TRAEGxHULL1 j GET 

			RICHT1xHULL1 j GET 

			RICHT2xHULL1 j GET 

			NORMxHULL1 j GET 

			\-> SCHNITTxPKT TRAEGxHULL RICHTxHULLx1 RICHTxHULLx2 NORMxHULL

                \<< SCHNITTxPKT TRAEGxHULL - DUP RICHTxHULLx2 CROSS 

					SWAP RICHTxHULLx1 CROSS 

					NORMxHULL DOT 

					SWAP NORMxHULL DOT 

					NORMxHULL NORMxHULL DOT 

					DUP ROT SWAP / UNROT NEG /

                \>> \-> KOEFFxRICHTx1 KOEFFxRICHTx2

                \<< KOEFFxRICHTx1 DUP

                  IF 0, \>= SWAP 1, \<= AND

                  THEN KOEFFxRICHTx2 DUP

                    IF 0, \>= SWAP 1, KOEFFxRICHTx1 - \<= AND

                    THEN j 2 \->LIST SCHNITTxHULL1 SWAP + 'SCHNITTxHULL1' STO

                    ELSE DROP

                    END

                  ELSE DROP

                  END

                \>>

              ELSE DROP

              END

            ELSE DROP DROP

            END

          NEXT

        NEXT SCHNITTxHULL1 DUP SIZE 1 - SCHNITTxHULL1 SWAP GET SWAP 1 GET

        IF ==

        THEN SCHNITTxHULL1 BODY BODY 'SCHNITTxHULL1' STO

        END SCHNITTxHULL2 'U1' STO SCHNITTxHULL1 'U2' STO

      \>>

    \>>

  \>>

\>>

So you get what you want (YGWYW ;-))

The input are four lists with 3-D vectors, the first two list discribe f. example the blue cylinder (there are only the coordinates needed in form of the bottom and top circle (the frames or formers of the cylinder)), the next two lists describe f. example the black cylinder (in the same form).

The user decides how fine (how many intersections points should be find), by the size of the lists. The lists of the different shapes can have a different size, but the pair itself must have the same size.

The user gets after some time two lists back "U1" and "U2", these two lists contain the intersections points in 3-D vectors.

It is not a analytical solution, it's numerical, but it works for all different shapes, which could be generated by two frames (two 3-D polygons).

It is a very clumsy piece of program, but you have to imagine, the calculator has to find out, if there are common points and where they are.

Greetings
peacecalc

[Pier Aiello]

Whoa, it's huge! Thanks for your contribution.

[Thomas Klemm]

Black cylinder

Assume r = 1:

x2 + y2 = 1

Blue cylinder

• P ... point on cylinder
  
  
• e ... direction of axis
  

Assume r = 1:

    | x |      | 0 |

P = | y |, e = | 1 |

    | z |      | 1 |

| P x e | = 1 * | e |

| x |   | 0 |   | y - z |

| y | x | 1 | = |   - x |

| z |   | 1 |   |     x |

2x2 + (y - z)2 = 2

(y - z)2 = 2 - 2x2 = 2y2

y - z = +/- sqrt(2)y

z = (1 +/- sqrt(2))y

So we end up with the following possible parametrisation:

x = cos(t)

y = sin(t)

z = (1 + sqrt(2))sin(t)

We could have expected this. For symmetric reasons the intersection curve must be in a plane. Cut that with a cylinder and you get an ellipsis.

Furthermore tan(67.5) = 1 + sqrt(2) and tan(-22.5) = 1 - sqrt(2).

Quote:

---

Where in this do you need a calculator?

---

Cheers

Thomas

[Pier Aiello]

Nice! Quick and elegant! (moreover i'll analyze it more asap)

About "where in this do you need a calculator?", it's not mandatory but if someone want to use it, why not?

[Thomas Klemm]

Oops, just noted that e should rather be (0, 1, -1). But it doesn't change much: just switch the sign of z.

Cheers

Thomas

[Csaba Tizedes (Hungary)]

Maybe it is required some thinking to calculate intersection curves, but the manufacturing is VERY simple, because the equal diameter pipes intersection curve is always two line in side view. This is a simple straight cut on pipes.

[Pier Aiello]

Why do you get a sort of triangle on the blue cylinder?

Remember that one "side" of the blue cylinder is untouched (the green one in the picture), while the other can be "cut" by the intersection line.

[Csaba Tizedes (Hungary)]

Those triangles are not visible, only I don't erase them.
I do not understand exactly the "one side is untouched"?!?

[Pier Aiello]

A cylinder has no sides :P, what I meant is the green (dotted) line/side of the blue cylinder in the first picture.

[Kimberly Thompson]

Pier

This is not an answer to your challenge, but rather an edification resource;

Link to Unwrapping Curves from Cylinders and Cones

the Pdf is highly enlightening.

BEST!

SlideRule

[Pier Aiello]

Thanks a lot :)