intersection point of two askew cylinders 1

[Thunderbird336]

Hello all, I have a family of parts that typically has two cylinders which intersect; I am looking for the point that is farthest from the midpoint of the largest cylinder's axis...

The angle between the two cylinders is either 45° or 60°
One cylinder, let's call it the main cylinder, is something like Ø20" and 30" long
The smaller one is something like Ø3" and arbitrarily, the same length
The smaller cylinder is translated up something like 6" and fore or aft something like 2".

We have a very small number of CAD licenses here so I am often unable to find this number when I need it to write a CNC program, what I need the number for is a starting point from which to mill a pocket. The work piece is set upon a rotary table with the midpoint of the larger cylinder's axis coincident with the rotary table center. The rotary table is then rotated to align the small cylinder's axis parallel with the horizontal machining center's spindle, or Z axis. The Y axis represents the upward translation (from the large cylinder axis) and the X axis represents the lateral translation (from the large cylinder's axial midpoint, same as rotary table center) of the small cylinder. It may be much simpler to do this in two steps though, figure out the intersection point before translating the small cylinder laterally; then I can easily use right angle trig to calculate the X and Z offset. I made two sketches of this in jpg format, the full solid and a cross section, I have (attempted) to attach them.

I know some Calculus, but not nearly enough to figure this out; it's beyond something that I can calculate with the kind of trigonometry that I know, it seems like spherical calculations. I have been unable to even find a good solid example of the standard form for a cylinder. I know what the standard form of a circle is and can use that for finding the intersection point(s) of two circles and hoped to find something similar for a cylinder. It would be helpful if I could program a calculator or make an Excel spreadsheet to get this coordinate for my programs for the (frequent) times when I need this point and cannot obtain a network license for our 3D software.

Thanks in advance,
-Gary

 

[ivymike]

"I am looking for the point that is farthest from the midpoint of the largest cylinder's axis..."

Okay, bearing in mind that all the points of intersection are the same distance from the axis of the larger cylinder (since the surface of that cylinder is all at the same distance from the axis), you need to find the point which is axially farthest from the midpoint of the axis.

Do the axes intersect?

I think you're looking then for the projected distance from the intersection of the axis of the second cylinder to the farthest point on the intersection of the second cylinder with the first. Given the angle between the axes and the radius of the second cylinder, you can find the projected axial distance (in terms of the large cylinder) from the axis of the second cylinder to its farthest intersection point. I think that point will always fall on a line on the surface of the large cylinder, parallel to the axis of the large cylinder, the projected length away from the intersection of that line and the axis of the second cylinder.

So
[ul]
[li]calculate [projected distance][/li]
[li]find where the axis of the second cylinder hits the surface of the first[/li]
[li]draw a line on the surface of the big cylinder parallel to its axis and passing through the axis of the second cylinder[/li]
[li]move ahead by [projected distance] to get to the point you want[/li]
[/ul]

 

[Occupant]

I am not clear as to what point you are looking for. Is it the intersection of the centers of the two cylinders in relationship to the center of the rotary table or what?

 

[ivymike]

(if your axes are not intersecting then the problem is much harder)

 

[MikeHalloran]

I can do trig, but it makes my head hurt.

I used to make paper patterns for producing workpieces like yours from rolled sheet, and for cutting intersections in them, using Rhinoceros, which is inexpensive as CAD systems go.

In the foreground of the attached photo is a 1:10 scale model cut from a quick checkprint of the patterns for the workpiece in the background.

My point is that with practice, it may be faster to generate a 3D model in Rhino and measure whatever dimension you want than to try and figure out even half the trig you will need.



Mike Halloran
Pembroke Pines, FL, USA

 

[Thunderbird336]

I am not clear as to what point you are looking for. Is it the intersection of the centers of the two cylinders in relationship to the center of the rotary table or what?

Do the axes intersect?

No, they do not intersect, that is why I included the qualifier "the farthest point" and that is precisely what is making this so tough to figure out. Since the small cylinder intersects at an angle and the system will be aligned to to the small cylinder's axis. The result of that is as you move laterally the distance from the large cylinder's axial midpoint (which is coincident with the rotary table center) changes, either larger or smaller depending on which direction we move in the lateral direction. The third complication is that as we move vertically, the number from the midpoint gets smaller due to the radius of the large cylinder, it would, of course, be largest if the small cylinder's axis intersected the large cylinder's axis and the least when the small cylinder's axis is tangent to the top surface of the large cylinder.

Were you able to view the two jpg files? The point I am looking for is the vertex of the acute angle between the two cylinders.

Thanks for looking at this...

 

[rb1957]

as i understand it the small pipe intersects the larger one, but not in-plane with the diameter of the larger pipe, yes ?

i got this idea from your 2nd pic (sectioning the smaller pipe) ...
you know the equation of the axis of the smaller pipe,
you know the equation of the inner edge (of the section), you know, 1 radius off-set from the axis
you know the equation of the edge (on the section of the larger pipe
your point is the interection of these two lines ... yes?

Quando Omni Flunkus Moritati

 

[berkshire]

What are you trying to do with this information? are you calculating a milling path to cut out the shape of the hole at the intersection?
You can calculate the points of intersection of the two tubes using the parallel projection method. Since it would appear that the centerlines of the two tubes do not meet, you may have to project an auxilary view to get your results. Sometimes referred to as double projection.
B.E.

 

[hydtools]

I used to do this on the drafting board projecting intersections of shapes. You could do it that way. How accurate do you need to be?

Ted

 

[Thunderbird336]

as i understand it the small pipe intersects the larger one, but not in-plane with the diameter of the larger pipe, yes ?

i got this idea from your 2nd pic (sectioning the smaller pipe) ...
you know the equation of the axis of the smaller pipe,
you know the equation of the inner edge (of the section), you know, 1 radius off-set from the axis
you know the equation of the edge (on the section of the larger pipe
your point is the intersection of these two lines ... yes?

that is in the ballpark but using the full radius as an offset gets me too far out; conversely, using the centerline of the small cylinder is not far enough out.

What are you trying to do with this information? are you calculating a milling path to cut out the shape of the hole at the intersection?

ultimately, a pocket gets milled, it's pretty much a 'spotface' for lack of a better term. this number that I am looking for is the starting point in the Z axis to mill the pocket from...

I used to do this on the drafting board projecting intersections of shapes. You could do it that way. How accurate do you need to be?

I may have to use something like that method in AutoCAD...

I was really hoping that there would be an equation for a cylinder, then some kind of system of equations could be setup and then solve for the point of intersection that is the furthest away from some known point like the midpoint of the large cylinder's axis but I can't begin to figure that out.

Perhaps it could be simplified though... in reality, the point I am looking for is the intersection of a circular section of each cylinder, right? The two circles are just not in the same plane, but the angle of the planes is known. I'm still not nearly smart enough to figure it out, but that at least may be a simpler problem. The difficulty still exists that there are many points of intersection, I wouldn't begin to know how to solve for the largest value (furthest one out) even if I knew how to put the two circles that lie in different planes in standard form.

 

[ivymike]

"The difficulty still exists that there are many points of intersection, I wouldn't begin to know how to solve for the largest value (furthest one out) even if I knew how to put the two circles that lie in different planes in standard form."


Well, if you can calculate and list 1000 of those points, then it is simple in excel to calculate the distance from another known point to each of those points...

 

[Occupant]

I am still not clear as to what you want - maybe my attachment sheds some light on it. You have a 2D sketch where you see the two Cylinders - please note, the small cylinder lies below the rotary table center in the top view, the rotary table center is defined by the two crossing center lines. The second page shows the small cylinder subtracted from the large one after two rotations and you are looking directly at the top of the hole that is left - again note: the rotations were both done about the center of the rotary table.

 

[rb1957]

your 2nd pic was a section thru the diameter of the small pipe, yes? so the lowest line of the small pipe is 1 radius from the center-line.

you also know where the large cylinder is being cut ... i'm assuming the small pipe axis is in the XY plane ... but if the small pipe is generally inclined then it harder to see how this plane intersects the large cyclinder.

Quando Omni Flunkus Moritati

 

[Thunderbird336]

your 2nd pic was a section thru the diameter of the small pipe, yes? so the lowest line of the small pipe is 1 radius from the center-line.

you also know where the large cylinder is being cut ... i'm assuming the small pipe axis is in the XY plane ... but if the small pipe is generally inclined then it harder to see how this plane intersects the large cyclinder.

yes, the second picture is a section thru the small cylinder, and both the large and small cylinders are in the XY plane...

 

[chicopee]

This problem is relatively easy to figure out with autocad solid. You draw the two cylinders with the solid commands; intersect the two solid cylinders in the proper orientations and positions; properly position the UCS in relation to the two intersected solid cylinders; subtract small cylinder from big cylinder; what you'll have left is the big cylinder with two holes; do a listing of both holes and you'll get the coordinates in terms of X, Y and Z values for your CNC.

 

[Cockroach]

In practice, I would solve it mathematically for the specific equations and orientation of each cylinder in 3D space. Problem is beyond the scope for this forum, you need to use multi-variable calculus and manifold theory in order to get what you are after. I suggest going to a mathematical forum for additional help. The intersection will be an ellipse whose eccentricity is the sine of axis inclination whith the first cylinder.

Historically, the mathematics develops from the "bird cage problem", which you can research on the net.

Regards,
Cockroach

 

[Sparweb]

It sounds to me like you are securing the large cylinder in the rotary table, but not actually using the rotation of the table to position and orient the part - it's just a convenient way to chuck the big round piece. So maybe this set-up is making it difficult to make subsequent set-up measurements, and that's what's getting in your way, not so much the mathematics.

I'm picturing a large cylinder in that rotary table, and the axis of the big cylinder is now pointed horizontally. The rotary table is clamped onto the milling machine's bed so that the cylinder points at a 45 degree angle to the bed's travel. Then you can come along with an end-mill cutter in the drive, ready to plunge into the big cylinder somewhere on the side. Except that you can't tell how far along the length of the cylinder you really are, because the starting point is somewhere on the slanted end of the cylinder. You may not even have enough bed travel to use a center point in the drive chuck to first find the end, and then traverse to the point of interest on the side of the cylinder. Tricky set-up, but there's probably a way to set up your tools to find the right point.

Have I described your situation correctly? I'd rather stop and take a breath at this point - no sense solving a problem you don't have.

PS your drafting team is supposed to be providing you with the information you need to make everything they draw. Tsk Tsk.


STF

 

[rb1957]

if both cylinder axes are in XY planes, offset from one another, i think it should be easy to solve.

your pic 2 is a section thru the small pipe diameter, ie the small pipe's XY plane. This is offset from the large pipes diameter plane by a distance,k; so that the two edges of the large cyclinder in pic 2 are 2*sqrt(R^2-k^2) apart.

that is, it should be easy to derive the equations of the two lines, the edge of the large cylinder and the edge of the small one.

Quando Omni Flunkus Moritati

 

[cowski]

Are you looking for something like this?

http://www.hsu.edu/uploadedFiles/Faculty/Academic_Forum/2011-2012/lloyd.pdf

www.nxjournaling.com

 

[IDS]

In practice, I would solve it mathematically for the specific equations and orientation of each cylinder in 3D space. Problem is beyond the scope for this forum, you need to use multi-variable calculus and manifold theory in order to get what you are after. I suggest going to a mathematical forum for additional help. The intersection will be an ellipse whose eccentricity is the sine of axis inclination whith the first cylinder.

It's really not that hard!

If both pipes are horizontal, you can treat the centre line of the small tube as the X axis, you then have lines at +- Rs at the edge of the tube in the XY plane, where Rs is the radius of the small tube. You know the vertical offset from the CL of the large tube to the CL of the small tube, so you can calculate the Chord length of the large tube section at the level where it intersects the XY plane. You know the position and orientation of the large tube centre line, so you can find the two lines in the XY plane at the ends of the chord. You now have a simple 2D line intersection problem.

A similar approach will give the intersection points for any level above or below the XY plane.

It gets a bit more complicated if one or both tubes are not horizontal, but there is a single plane defined by the CL of the small tube, the IP with the offset point of the large tube, and the line through this point parallel to the large tube CL. If that plane can be found, you can rotate the coordinate system so that the CL of the small tube lies on the X axis and the CL of the large tube is parallel to the XY plane, then proceed as above.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/