Thread: Calibrating a Surface Plate

Just a thought before you start lapping - is it worth checking (or playing with) the support points. I can never remember the proper name for them (Airey points?) but if you have a bulge in the middle of the plate it may be because the ends of the plate are sagging, a result of the supports being too close together. Alternately even if they are properly placed you may be able to reduce the bulge by moving them out a little bit.
If you are chasing micro inches that may be all that is needed to even things out. You may be able to claim that at least part of the table is AA.

Michael

Yes Airy points Michael, Bessel points is the other one. The former distance between the points is 0.5577 of the overall length and minimises deflection of a horizontally supported beam. The latter is very similar, 0.559, and minimises change in overall length. Technically I believe they apply to beam support and not a plane, and since the support is on 3 pads, then the odd pad can't possibly lie in an Airy position, but it's certainly what we've come to expect from surface plate support. I'd expect a quality plate like that to have the support pads fitted when manufactured, but I think yours is a good idea. Try simple, reversible, solutions first. Tweaking them slightly could well bring up a low area without dropping the other areas so much that they then fall out of spec.

Pete

Originally Posted by RayG

The things you do when you're easily amused..

Surface Plate Effects - YouTube

Regards
Ray

If all else fails you could turn it into an air hockey table....

I would say grow up Josh, but let's face it, I'd be doing the same thing.

Hi Michael, Pete,

Interesting idea, I wonder how much it would move by shifting the supports? Currently it's supported on 3 rubber pads, and the frame looks like it's original to the plate. there are additional rubber pads on the corners, but these don't touch the frame. Josh machined up some ball bearing mounts in preparation for remounting this plate, but on closer inspection when we got it home we decided that there was nothing wrong with the factory original 3 point supports. ( Aside from the fact that we would need to to muck about getting hydraulic jacks in place to lift the plate, lifting this thing is a non-trivial exercise.. )


Hi Ewan,

Air hockey came to mind.. the ultimate in backyard pool tables was another idea..

I also found a youtube video of Nick Mueller doing the same thing.. Metal scraping: Surface plate, will it hover? - YouTube
At least we are in good company...

Here is a homework question, what is the curvature of the earth over the length of a 6ft surface plate, answer in microinches..

Regards
Ray

Hi Ray,
If you where on the equator at sea level, the earths radius is 6378.137km or so. Over 1800mm that equates to .063896um....I think I got the decimal place in the right spot anyway.

A question for you though, is the laser effected by gravity, and if so how much? Will its beam follow the curvature of the earth or stay on a true straight line.

On to the serious stuff, my one concern with the pool table would be the accuracy of the balls, I have heard that if a pool ball was the size of earth it would have hills and valleys about twice the size of Everest and the laurentian abyss. Not nearly good enough for the table.....

0.000243126521"

Greg

Originally Posted by Greg Q

0.000243126521"

Greg

I get 0.0000025928"

Stuart

p.s. I'd thought of asking you to turn the plate around and recheck it. just to see how much difference it makes....... though maybe thats a job for a smaller plate

Originally Posted by Ueee

A question for you though, is the laser effected by gravity, and if so how much? Will its beam follow the curvature of the earth or stay on a true straight line.

Yes, but nothing like the curvature of the earth(unless you have a black hole handy)
Even in Rays it couldnt be much.

What radius are you working it on. I get 2.51559uin from my um measurement. I missed the uin bit in the question.

251106299 inches

I should add it wouldnt surprise me at all to find that I was wrong, as I am just punching numbers into this cal

Circle Calculator

Stuart

I was quickly interpolating from a larger scale calculation...something you cannot do as obviously the circle falls from the tangent in a non linear fashion. Two tenths over six feet doesn't even smell right...I believe that two millionths is closer to the mark.

Inches, btw

Greg

That agrees with Josh's 2.5 uinch calculation, and no you can't scale from a larger length, ( that's what I did using 8" per mile, gives you crazy numbers.. )

Anyway, the reason for the question, is related to a little side project I'm working on when I have time... the brief concept is that MEMS sensors. ( microscopic silicon structures) have got differential inclinometers down to around the 0.001 degree level, ( 3.6 arc-seconds) which works out to 0.02 mm/meter order of sensitivity, these sensors cost around the $50-$100, but with oversampling and a bit of fancy signal processing footwork, I think I can get under the 1 arc-second sensitivity... so a differential level setup which is sensitive enough to calibrate surface plates could be an affordable project...

The chip I've been looking at is this one SCA103T Inclinometers | Murata Electronics Oy
data sheet is here http://www.muratamems.fi/sites/defau...261700a3_0.pdf

Regards
Ray

Geez Stu, and here was me scribbling on the envelope my phone bill came in today. Gotta love the little black book and its sine tables though, well, at least I thought so until it clicked that the angles are so close to 0 and 90deg its not funny.....

I'd love to here more Ray, I'll check out the link when I'm on the pc tonight.

The answer depends a little on how you construct your model, No matter which way you do the calculation though it is relatively insensitive to the precise radius given for the earth. It is however very sensitive to the dimensions given for the surface plate. ie. nearest 10km for earth but nearest cm for plate or miles and inches if you prefer. I uesd 6370000m and 1.82m

If constructed so that the centre of the plate is at a tangent; the difference in length to the centre of the earth at the ends of the table compared to the centre of the table is about 2.57µin

If constructed so that one end of the plate is at a tangent; the difference in length to the centre of the earth from each end is about 10.4µin. This would be equivalent to looking out a mile and measuring a 8 inch drop.

Originally Posted by Brobdingnagian

If constructed so that the centre of the plate is at a tangent; the difference in length to the centre of the earth at the ends of the table compared to the centre of the table is about 2.57µin

That's almost how I did it, but I made the table a chord and then worked out the triangle made by half the chord and the radii as the hypotenuse. The difference between the opposing side and the radius was my answer.....