intersection point of two askew cylinders
intersection point of two askew cylinders
(OP)
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
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





RE: intersection point of two askew cylinders
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
RE: intersection point of two askew cylinders
RE: intersection point of two askew cylinders
RE: intersection point of two askew cylinders
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
RE: intersection point of two askew cylinders
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...
RE: intersection point of two askew cylinders
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
RE: intersection point of two askew cylinders
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.
RE: intersection point of two askew cylinders
Ted
RE: intersection point of two askew cylinders
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.
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 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.
RE: intersection point of two askew cylinders
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...
RE: intersection point of two askew cylinders
RE: intersection point of two askew cylinders
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
RE: intersection point of two askew cylinders
yes, the second picture is a section thru the small cylinder, and both the large and small cylinders are in the XY plane...
RE: intersection point of two askew cylinders
RE: intersection point of two askew cylinders
Historically, the mathematics develops from the "bird cage problem", which you can research on the net.
Regards,
Cockroach
RE: intersection point of two askew cylinders
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
RE: intersection point of two askew cylinders
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
RE: intersection point of two askew cylinders
http://www.hsu.edu/uploadedFiles/Faculty/Academic_...
www.nxjournaling.com
RE: intersection point of two askew cylinders
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/
RE: intersection point of two askew cylinders
I've done it on the board, and I've done it in AutoCAD, and Rhino is an order of magnitude easier and faster. Solidworks can probably do it too, but Rhino is faster and much cheaper.
Mike Halloran
Pembroke Pines, FL, USA
RE: intersection point of two askew cylinders
To be noted that the farthest point is not in the plane of the section shown by Thunderbird336 and of course that the intersection is not an ellipse.
prex
http://www.xcalcs.com : Online engineering calculations
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RE: intersection point of two askew cylinders
Input (grey shaded on the Cylinders sheet) is the radius and length of the cylinders. vertical offset of the small cylinder, and longitudinal offset of the IP with the CL, and the angle. You can then adjust the level at which you want the intersection coordinates.
The XY coordinates are displayed in blue, and plotted on a horizontal plane, similar to your second sketch in the original post.
Note that I haven't checked anything.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: intersection point of two askew cylinders
prex, seems like you understand this the best, it really is not simple at all. You can get in the ballpark figuring in a plane, but it is still pretty far off, like 1/2" or so depending upon how far up the larger cylinder the smaller one intersects...
This is a similar exposition...
Link
I am just not sharp enough to put them to use, he makes mention of Cartesian equation...
A curve in the xy-plane has an implicit Cartesian equation of the form
m(x, y) = 0.
and I don't even know what that is, so I cannot get there from here... but I will look over the article that cowski linked. Thanks all for so much help on this.
RE: intersection point of two askew cylinders
The spreadsheet looks pretty good; I only verified the one example case in my CAD software. I think the spreadsheet with the use of Excel's solver add-in will get the points of interest. If I may suggest one (unnecessary) change, it would be to list the valid range for the IP Z coordinate input. It can range from offset 1 - small cylinder radius to offset 1 + small cylinder radius (unless the offset + the small radius > the large radius). This would give the user an indication of valid values for input and would also let you add constraints to the input for the solver (works without them, but would help capture intent).
Thunderbird336 (and others who are interested),
You can use Excel's solver to search for a minimum value of Small 1 Y (cell C22), by changing the value in IP Z coordinate (cell B14), then search for a maximum value of Small 2 Y (cell E22) by changing the value in IP Z coordinate (cell B14). I've only verified 1 test case, but I think the strategy used in the spreadsheet is sound.
www.nxjournaling.com
RE: intersection point of two askew cylinders
I will check this out tomorrow, I apologize... I did not see this until I got home and read cowski's reply to you.
Gary
RE: intersection point of two askew cylinders
The spreadsheet calculates the coordinates of the intersection points on any specified plane, so you can adjust the level of the plane, using the Excel solver, to maximise or minimise whatever you want.
I have added a calculation of the distance of the intersection points from the mid-point of the large cylinder axis, and set up the solver to maximise the larger dimension.
If that's not what you wanted, please let me know.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: intersection point of two askew cylinders
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: intersection point of two askew cylinders
That is an amazing spreadsheet! I am often astounded by what people can do with Excel.
The calculation seems to break down once the small cylinder is translated up past the point of intersection with the edge of the large cylinder... I attached a screen shot of the one I am currently working on. Just for clarification, as I said before, when I can get a CAD license (such as now!) I can get these numbers out of our 3D software. Actually, to be really honest, I only know how to get close even with it, but once all of the design engineers arrive, I have no hope of getting a software license until the next day. (Just in case your wondering why I want to calculate it apart from 3D CAD software)
Is it possible to tweak the calculation to allow for the small cylinder to go higher, as in my picture? If you cannot see the picture I will upload it to the Engineering.com server, it's on Dropshots presently and, if I did it properly, is in a public folder.
Thanks so much!
-Gary
RE: intersection point of two askew cylinders
Quando Omni Flunkus Moritati
RE: intersection point of two askew cylinders
www.nxjournaling.com
RE: intersection point of two askew cylinders
I can only get a CAD license early in the morning before the design engineers arrive as I said in my post to Doug...
No, cowski, this is the same problem, I thought it would be easier to describe as a cylinder, after all, the fillet edged of the pocket is a cylinder in fact and it will always be the furthest point out where the two cylinders intersect, so it is all that matters in this case.
-Gary
RE: intersection point of two askew cylinders
Wouldn't this just be changing the small cylinder z offset? Or am I missing something?
www.nxjournaling.com
RE: intersection point of two askew cylinders
As I said to Doug (above) "The calculation seems to break down once the small cylinder is translated up past the point of intersection with the edge of the large cylinder..." the spreadsheet will not give a result once the small cylinder goes high enough that the point of intersection is "inside" the large cylinder.
I put some numbers in the spreadsheet and kept moving up on the Z, this is as far as I can get...
Small Large
Cylinder radius 1.5 16.875
Length 40 40
Offset1 (Z small cylinder CL) 4.284
Offset2 Y IP on CL 2
Angle to Y axis -30 0 degrees
beyond this, it will give an error, the same thing will happen if you shorten the small cylinder to, say, 30"
There is still a point of intersection, it is just no longer at the edge.
-Gary
RE: intersection point of two askew cylinders
www.nxjournaling.com
RE: intersection point of two askew cylinders
RE: intersection point of two askew cylinders
Very cool! I actually forgot that AutoCAD can do 3D, I have seen it done and I have walked through a couple of tutorials but it's been a few years. I have never been denied an AutoCAD license, so if I can figure it out, that would work, thanks for looking at this and helping me see that I do have resources available to do this in.
-Gary