Geometry question

[88keys]

Im looking to test the accuracy of measuring hole diameters using bearings and measuring how deep it falls into the hole.

So if I have a .200 diameter bearing (sphere with a .100 radius) and when inside the hole, the distance from the hole surface to the top of the bearing measures .163; there should be a way to calculate the hole diameter by determining the diameter the sphere in at that point (.163)

I know there is a formula to do this but I can't remember it to save my life.

 

[MongGrel]

That really has no meaning unless it is something like a taper reamed hole, and the sides have a specific angle.

All you get from the bearing would be that it is .263 to the center of it from the surface when it is out of conformance.

If you are measuring a tapered hole the angle would be the omitted factor.

Other than that, it just means your hole is not .200 at that depth.

Or it could be larger and oblong, if QA is involved it would depend on ISO specs and GMT.

That really does not involve a formula.

 

[DrPizza]

I think I've figured out what you're trying to say. But what you did say implies that you think by measuring how far down a bearing goes, you can measure the diameter of the hole.

measuring how deep it falls into the hole....

I can take a 1" drill bit and drill a hole 6" deep or 12" deep. The bearing is going to fall a lot further in the 12" deep hole, and have nothing to do with the diameter.


What I *think* you mean is that the bearing sits on the hole and doesn't fit down into it. There may be an easy/easier formula, but it's not something I've ever used. What you can do is use the formula for a sphere, then find a formula for the trace (cross section) of the sphere at any point. x^2+y^2+z^2=r^2. If your radius is 0.1, and the top of the sphere is .163 units above the surface, then you're looking for the formula for the circle at a value when z=-0.63 (The surface is 0.63 units below the center of the sphere.) So, plug -0.63 into the equation for z, plug in 0.1 for r, and do a little arithmetic so that it says x^2+y^2 = some number. The square root of that number is the radius of the hole.


Alternatively, if your bearing is very hard, you could smack it really hard with a hammer a few times. Then, the radius of your hole is 0.1.

 

[Carson Dyle]

What I *think* you mean is that the bearing sits on the hole and doesn't fit down into it.

Seemed clear to me.

 

[MongGrel]

I think I may have misunderstood the original question a bit myself, my bad.

In a practical physical application you could just measure it other ways, I was thinking the bearing was actually being dropped into a hole. We even had test setups that would just register airflow around a bearing in the past for really tight tolerances.

 

[MongGrel]

Seemed clear to me

so your answer is ?

 

[Carson Dyle]

so your answer is ?

The problem proposed was clear enough.

 

[MongGrel]

The problem proposed was clear enough.

So you still have no answer I guess, as obvious as the question was ?

It's pretty trivial these days if you have CAD/CAM software and just drop a 3D sphere on top of a hole these days. Most engineers are not messing with slide rules much these days either.

 

[Carson Dyle]

DrPizza gave the answer. It's just x = sqrt(r^2 - y^2), where y is the distance of the centerline of the sphere above the hole.

Or using the height (h) of the bearing measured above the hole: x = sqrt(r^2 - (h-r)^2)

 

[PottedMeat]

bearing + height gage?

comparing the accuracy to this + micrometer?

 

[DrPizza]

Seemed clear to me.

You must be operating with a different definition for "into" than we do.

 

[Carson Dyle]

You must be operating with a different definition for "into" than we do.

Because exactly like you said - you drop the ball "into" the hole and it tells you nothing about the hole's diameter. It was pretty clear. But whatever.

 

[Fenixgoon]

from a slightly more practical standpoint, not only are the dimensions of your bearing and hole important, but also the form - how spherical is your bearing compared to a perfect sphere? how cylindrical is your hole compared to a perfect cylinder? real parts have tolerances, and a mismatch between the sphericity of the bearing and the cylindricity of your hole could mean that the bearing won't go down the hole, even if the cross-sectional clearance were ok at any given point.

but on the math side, Dr. Pizza and carson dyle have it.

also, you would not using bearings to check hole depth (i thought we were talking hole diameter). just buy a cheap caliper and have at it or using a proper hole depth measuring device

 

[Fenixgoon]

nvm