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[machinist] Trig Solution to Tapering Hole Diameter Problem
- From: Lawrence London <lfljvenaura@gmail.com>
- To: machinist@lists.ibiblio.org
- Subject: [machinist] Trig Solution to Tapering Hole Diameter Problem
- Date: Sun, 29 Dec 2013 20:23:05 -0500
http://www.practicalmachinist.com/vb/general/trig-solution-tapering-hole-diameter-problem-277573/
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Trig Solution to Tapering Hole Diameter Problem
For anyone interested in the question I am posting the trig solution to the problem of spacing holes on a tapered arm with uniform edge-to-edge spacing along the axis of their centers. I decided to start a new thread as this trig problem was not really the main point of the first thread Is there a scratch-free layout method available? . The trig solution included is applicable to any number of holes with any taper and any spacing.
The "bottom line" solution is in red at the bottom of the page. For completeness the derivation of the solution is included above that.
There are two images attached. One is a photo of the drawing of just two circles with an exaggerated taper angle as this setup allowed a more clear separation of some of the tangent and intersecting lines than would have been possible if the circles were drawn to the same taper as the part I was making. The second picture shows the part I intend to make drawn to scale with the spacing of the holes and their radii indicated.
The work I did on this makes no sense in some ways as a nearly as good and practical result could have been obtained in a much shorter time by just using paper and pencil or a CAD program and empirically fitting the circles between the lines drawn. I guess the challenge of getting a math solution sucked me in.... Maybe this math solution will be of use to someone else who doesn't care to grind through the mental gymnastics to figure it out.
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The goal of this exercise to determine a method to calculate the diameters of circles of smoothly diminishing diameters and spaced at a specified distance from circumference to circumference and with their centers falling on a straight line. In the diagram shown only two circles are considered. But, once the calculation is made for one pair of circles, subsequent calculations can be made using the same formula.
Definitions:
a is the angle subtended by one line tangential to both of the circles and another center line through the center of the two circles.
R is the known radius of the larger circle.
S is the distance between the circles as measured on a line connecting their centers. Its length is arbitrary but is constant.
r is the radius of the smaller circle. Its value is unknown and we want to calculate it relative to R for a given S.
First we need to find the height of the line F-E:
A-B = R/cos(a)
E-D = (S+R)
D-B = [E-D/cos(a)] sin(a) = (S+R) tan(a)
A-D = A-B - D-B = F-E = [R/cos(a)] - [(S+R) tan(a)]
F-E = [R/cos(a)] - [(S+R) tan(a)]
Now that we have calculated F-E we can find the radius (r) of the smaller circle by determining F-E in terms of r:
H-G = H-J + J-G = r sin(a) + J-G = r sin(a) + r = r[sin(a) + 1]
H-E = H-G/cos(a)
E-G = H-E sin(a) = {r[sin(a) + 1] sin(a)}/cos(a) = r[sin(a) + 1]tan(a) correction to original post in bold below
F-E = E-G + rcos(a) = r([sin(a) + 1]tan(a) + cos(a))
So we have now determined F-E in terms of R and S and also determinedF-E in terms of r.
So: F-E = [R/cos(a)] - [(S+R) tan(a)] = r([sin(a) + 1] tan(a) + cos(a))
r = {[R/cos(a)] - [(S+R) tan(a)]}/({[sin(a) + 1] sin(a)} + cos(a))
r= [R/cos(a)] - [(S+R) tan(a)]
.......{[sin(a) + 1] tan(a)} + 1 ........Correction Needed---see restated formula on next line per RayG---Thank You, RayG
r= [R/cos(a)] - [(S+R) tan(a)]
.......{[sin(a) + 1] tan(a)} +cos(a)
The restated formula above is correct (I think!)Last edited by dgfoster; 12-29-2013 at 12:42 PM. Reason: formatting
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12-29-2013, 12:11 AM #2
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All I can say, is I'll keep you DGfoster, for any wild math I need help with. I'm not to bad in trig, but I'm
Stuck on formulas for an equal angular spiral, and the formulas for transformation from log-polar to Cartesian coordinates. Also known as a logarithmic spiral. I'd like to have an understandable formula where I could spread sheet it and have it geometrically accurate where I could plug in an angle variable. I can build the excel sheet.
If thats ok,
Stan- -
Nice work dgfoster, here's another solution, I suspect it's equivalent to yours.
Once you have calculated the first r, the successive radii, can be calculated using the ratio r/R, and spacing between centers is r+S+R ( using the new r values each time of course.. )
Interesting problem, trickier than you'd first think.
Ray
Edit: A quick spreadsheet check, to compare the two formulae,
For R=2, S=0.25, angle 5 degrees.
Formula 1 ( dgfoster ) r= 1.654
Formula 2 ( this post) r= 1.659
I suspect rounding errors in the spreadsheet trig functions, ( no it's another problem... see next post.) -
Hi dgfoster,
Originally Posted by dgfoster
First we need to find the height of the line F-E:
A-B = R/cos(a)
E-D = (S+R)
D-B = [E-D/cos(a)] sin(a) = (S+R) tan(a)
A-D = A-B - D-B = F-E = [R/cos(a)] - [(S+R) tan(a)]
F-E = [R/cos(a)] - [(S+R) tan(a)]
Now that we have calculated F-E we can find the radius (r) of the smaller circle by determining F-E in terms of r:
H-G = H-J + J-G = r sin(a) + J-G = r sin(a) + r = r[sin(a) + 1]
H-E = H-G/cos(a)
E-G = H-E sin(a) = {r[sin(a) + 1] sin(a)}/cos(a) = r[sin(a) + 1]tan(a)
F-E = E-G + r = r([sin(a) + 1]tan(a) + 1)
So we have now determined F-E in terms of R and S and also determinedF-E in terms of r.
So: F-E = [R/cos(a)] - [(S+R) tan(a)] = r([sin(a) + 1] tan(a) + 1)
r = {[R/cos(a)] - [(S+R) tan(a)]}/({[sin(a) + 1] sin(a)} + 1)
r= [R/cos(a)] - [(S+R) tan(a)]
.......{[sin(a) + 1] tan(a)} + 1
There is an error, in your derivation of FE for the smaller circle,
The line...
F-E = E-G + r = r([sin(a) + 1]tan(a) + 1)
Should be F-E = E-G + rcos(a) = r([sin(a) + 1]tan(a) + cos(a))
This makes the corrected formula
[R/cos(a)] - [(S+R) tan(a)]r = --------------------------------------------{[sin(a) + 1] tan(a)} + cos(a)With that correction, I get same answers as my version.
Ray
PS... for small values of a cos(a) will be close to 1, so the error only becomes noticable at larger values of a.Last edited by RayG; 12-29-2013 at 07:54 PM. Reason: formatting attempt...
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12-29-2013, 12:03 PM #5
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Another math man, wonderful!
I do need some help, and am willing to contribute to get an accurate answer. I'll ask through Private Messaging when I've got it together and hope you (plural) are willing.
Regards,
Stan- -
RayG,
You are oh so correct in pointing out an error that slipped through in my derivation. I must have looked cross-eyed at the diagram as it is clear that the line segment F-G is indeed rcos(a) not r. I do appreciate your pointing out the error (and doing so kindly). You are also correct that, since the cos in my part drawing is so small that the error would barely show up. I corrected my original post.
I like your solution better than mine as it is a simpler way to look at the problem and thus less error prone and easier to understand.
I had hoped that, if there were errors in my solution, that they would be picked up by anyone who took the time to really check it.
Thank you.
I will re-run the numbers in my part drawing.
Denis -
I want to give props to dgfoster and RayG as well as the PM community....first for being smarter than I with trig and secondly and most importantly for being so respectful....I frequent a few other forums and a thread like this may might have turned into a who's penis is bigger thread.....good job guys
This is one of the many reasons I enjoy this forumLast edited by xdmp22; 12-29-2013 at 07:48 PM.
Robert Campbell Jr. likes this. -
Somewhere, if I've still got it, I've got an old Lotus spreadsheet that I made in about 1986 for doing about the same thing. Back in that life, I was building generators. These had a tapered input shaft. The inputs I had were the 2 ball sizes and the depth (or amount above) of the balls from the datum surface. Output was the gageline depth and angle.
I don't even know if I still have it, but it took a lot of de-bugging in Lotus. Boy, was that a long time ago! They didn't have a way to measure the parts before that. They just fit them to a crankshaft and checked for wobble. I was working on a special gage until we got a new plant manager.
JR -
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Hi Denis, Originally Posted by dgfoster
I had hoped that, if there were errors in my solution, that they would be picked up by anyone who took the time to really check it.
Thanks, it was an interesting problem
The only reason I found the problem, was, I couldn't see why there was a difference between the answers for the two methods.
Incidentally, your revised version can be re-arranged to give the same equation. I won't do it here, but if you multiply top and bottom by cos(a), and substitute sin(a)/cos(a) for tan(a), and use the identity sin2(a) + cos2(a) =1, you get the same equation I got.
Ray
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[machinist] Trig Solution to Tapering Hole Diameter Problem,
Lawrence London, 12/29/2013
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- [machinist] Trig Solution to Tapering Hole Diameter Problem, Lawrence London, 12/29/2013
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