[Metrology] Measuring An Internal Taper Using Two Balls. + Question

I can't see the spreadsheet. However:

The radii of the balls to the line of contact form a right angles with the wall of the cone: that it your right angle (not the angle at the ball center). The triangle formed by the radius of the larger ball to the line of contact, a line along the cone to the line of contact of the smaller ball, and a line from the line of contact of the smaller ball to the radius of the larger ball parallel to the center line is similar to the triangle formed by the center line, the radius of the larger ball, and the cone wall. The hypotenuse of this little triangle is the center to center distance of the balls because the aforementioned radii are parallel. The short side of this triangle is the difference in radii. Therefor the half-angle is the arcsine of the difference in radii divided by the center to center distance.

I'll upload a drawing later.
 
Tony Wells said:
There is one subtle factor I didn't see addressed in a quick look at your math. That's the effect of the contact line between the ball and the unknown taper NOT being on centerline of the ball. So, in reality, you cannot use the radius of the ball directly in your calculations. Since the bore is tapered, the contact line is below centerline of the ball, and hence the radius value must be somewhat less than the radius of the ball(s). The problem is that you don't know where exactly that contact line is, due to the taper, which is unknown. So it presents a somewhat circular calculation with an unsolved variable. The contact line defines a plane that lies on the tangent point between the taper and the ball.

The ball measurement does work reasonably well on shallow tapers, because that tangent point approaches the radius contact of the ball, but when it does, obviously you then have a cylinder, not a tapered bore. The steeper the taper, the more pronounced this factor becomes. I learned this the hard way a long time ago when making some really critical tapers and decided it would be slick to use 2 balls to measure it. Then the customer sent me a precision ground taper plug to do a blue fit. Took some thinking to see why the methods disagreed. This was before the Internet and all the information it offers, but I did have the Machinery Handbook. And some helpful engineers on the other side of the coin.

This has been discussed on a few other fora plus Googe Groups, etc.
Click to expand...

Thanks for the input,

I did address the taper touching the ball below the centerline of the ball in the ball height / mouth opening pages since the angle is known.
My first thought was to use this same method on the taper angle solution, but since the angle was not known the resulting equation required an iterative solution. where you put your first guess on angle into the equation and use the answer to fine tune your next guess, and repeat until you get the answer. It is kind of like the "I'm thinking of a number between 1 and 100 game" where you use the higher and lower responses to adjust your next guess. Excel does have a way to do iterative solutions, but I was looking for a clean solution that could be solved with a pocket calculator if needed.

After looking at that I decided to change the orientation of the problem, so my inputs would only be things that I knew.
The orientation for the other ball height solutions has the ball height measured on the Y axis, and the ball centerline on the X axis with the actual ball taper contact point below the centerline. I chose to rotate the problem so the taper wall on one side of the ball was the X axis, and the ball heights are measured on the hypotenuse. This orientation automatically makes the Y axis a line from the ball center down to where the ball touches the taper. This makes the centerline of the ball irrelevant at that point. If you shift the X axis up to the centerline of the small ball you now have an easily sovable right triangle where the side opposite to the angle (Y axis) is equal to the the large radius minus the small radius and the hypotenuse is the distance between the two ball centers.

I will make a video of the taper derivation tonight and will post it.

Here is a video I made of the derivation of the ball height vs mouthopening.
I tend to use a trigonometric solution instead of a geometric solution so the equations may look different than what some are used to seeing.

I'l post a video tonight,

Thanks,

Chris
 
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RJSakowski said:
The equations for solving the taper measurement by two balls are fairly complicated, at least for this poor soul who last took a trigonometry class in 1959. Here is a drawing of the geometry and the equations derived from it. The last equation gives the taper in terms of the diameters of the two balls and the distance between them as measured by Chris above. Microsoft Excel can be used to solve the equation using an iterative process which converges quickly. View attachment 136642

It works like this: d1, d2, and h are fairly straightforward. It is the cos(a) that is the problem. Assume that a = 0 and therefore cos(a) = 1. Substitute in the equation and solve for tan(a). Calculate a new value for a and cos(a) and go through the calculation again. The calculated value will soon converge on the true value, small angles faster than larger ones.

I will upload the Excel spreadsheet to the download section.

I should add that this is not the way that I usually calculate the angle. As Tom stated, I draw the geometry in a good drafting program (I use SolidWorks) and let the CAD program solve the problem.
Click to expand...

Thanks for the input.

I initially solved it in a similar fashion but was looking for a solution that didn't need an iterative solution. So I changed the orientation of the problem (see my response above that describes the new orientation.)

I will post a video of my derivation tonight.

Thanks,

Chris
 
John Hasler said:
I can't see the spreadsheet. However:

The radii of the balls to the line of contact form a right angles with the wall of the cone: that it your right angle (not the angle at the ball center). The triangle formed by the radius of the larger ball to the line of contact, a line along the cone to the line of contact of the smaller ball, and a line from the line of contact of the smaller ball to the radius of the larger ball parallel to the center line is similar to the triangle formed by the center line, the radius of the larger ball, and the cone wall. The hypotenuse of this little triangle is the center to center distance of the balls because the aforementioned radii are parallel. The short side of this triangle is the difference in radii. Therefor the half-angle is the arcsine of the difference in radii divided by the center to center distance.

I'll upload a drawing later.
Click to expand...

Sorry you cant see the spreadsheet, I can e-mail it to you if you like.

Based on what you wrote, I think you have the same solution as mine.

I will film and post a video of my solution tonight.

Thanks,

Chris
 
That's pretty much the same solution as the one I came up with. It is indeed very simple: arcsin((d1-d2)/2h) where d1 is the diameter of the large ball, d2 is the diameter of the small ball, and h is the center-to-center distance.

Did you also have Miss Grout for trigonometry?
 
John Hasler said:
That's pretty much the same solution as the one I came up with. It is indeed very simple: arcsin((d1-d2)/2h) where d1 is the diameter of the large ball, d2 is the diameter of the small ball, and h is the center-to-center distance.

Did you also have Miss Grout for trigonometry?
Click to expand...

No, I had Riad Daqqaq. And no, we could not pronounce his name. LOL
I met my future wife in trig class. We studied together for this class.
Chris
 
I uploaded the spreadsheet as a zip file in the Downloads, Calculators for Metal Working category. http://www.hobby-machinist.com/resources/socket-taper-calculation-by-two-ball-method.3098/

John Hasler said:
That's pretty much the same solution as the one I came up with. It is indeed very simple: arcsin((d1-d2)/2h) where d1 is the diameter of the large ball, d2 is the diameter of the small ball, and h is the center-to-center distance.
Click to expand...
John, I plugged the numbers from my SolidWorks drawing into your formula and it comes out with an angle of 5.4502º. I also checked the Excel spreadsheet against the SolidWorks drawing and the angles agree to eight decimal places so I'm quite sure that my equations are correct. (On the original drawing, the dimensions for d1, d2, and h were rounded off which I corrected in the drawing in the spreadsheet.)

I would be curious to see how you arrived at your equation.
 
Very good, gentlemen. That actually is the same method I used nearly 20 years ago. I didn't have a pc, nor excel, so it was all longhand (and a pain). I appreciate your work here and providing the spreadsheet for us. Thank you all.
 
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