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



#17


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.


#18

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?


#19

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.

#20

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


#21

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.


#22

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


#23

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

#24

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


#25

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?


#26

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

#27

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.


#28

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.


#29

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


#30

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

#31

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


#32

Thanks a lot :)


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