Justify my debt

You have a 14x44 mtal lathe. A mill. Nothing fancy. Someone tells you to turn a 1" brass bar EXACTLY 2 feet 2 inches long +- .001".

I just bought a 29" Starrett vernier.

I believe, that without this most simple, if expensive, tool, there is no cheaper way to fabricate a metal bar EXACTLY 2 feet 2 inches long.

Its food for thought.

The maximum length bar you can fabricate to an EXACT length is determined, most cost effectively, solely by the biggest vernier caliper you have.

i.e., for those of us who own an 8" dial caliper and a 6" micrometer at biggest, cannot create a metal bar of 9.700 inches +-.001"

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I would turn the brass bar to diameter a bit long, then clamp it in V-blocks on the mill table and use the DRO and a new end mill to mill the ends to length.

Or I'd buy one of those cal-x-tenders (sp?) which go onto regular calipers.

Or else I'd figure the cost of the 1" bar to include the Starrett vernier and quote the guy a shocker of a price and only buy the tool if he is willing to pay for it.


Ben wrote:

Reply to
Grant Erwin

And remember, at that length, it will grow/shrink .001 for each 4.5 degree F change in temperature.

Reply to
Jim Stewart

Of course you can. You stack the requisite number of smaller pieces, each exactly measured, end on end and compare the total length with a dial gauge to the piece you want to make. In the same way slip gauges are stacked together to make a length for which no single slip gauge exists in the box.

Dave Baker - Puma Race Engines

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I'm not at all sure why women like men. We're argumentative, childish, unsociable and extremely unappealing naked. I'm quite grateful they do though.

Reply to
Dave Baker

Find out what they are going to be checking your part with. If it's a tape measure, you are in the clear.

If they have a long master vernier caliper, ask to borrow it.

Or, turn it slightly over-long, and submit it. Then they will give it back and say, 'hey, dunderhead, it's

58 thousanths too long!'


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Reply to
jim rozen

Takes a bit of work but you can make extenders for any caliper or micrometer to measure longer items. With longer stuff, make sure that you control temp of the metal well or you will lose accuracy just from the expansion of the metal as noted elsewhere in this thread!

-- Bob May Losing weight is easy! If you ever want to lose weight, eat and drink less. Works evevery time it is tried!

Reply to
Bob May

Turn it close, within a few thou either way. If they complain, leave it outside in the cold for a few minutes if it's too long, or stick it down your trousers (don't do this while going to a bar for lunch or you will attract the wrong sort of new friend) if it's too short.

The real answer, I suspect, is ask them why they have such a silly spec. That, or justify and pay for a new fancy vernier...I've done worse!

If they have a sample, it's dead simple to make yours the same length as theirs, of course.

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Ok, I'll bite. What's the easiest way to compare the two, accurate to a few tenths?

Reply to
Jim Stewart

Make a fixture out of some angle iron, with a stop on the bottom, and a piece to hold a good dial gage at the top. Lift the indicator button, put in your standard, and zero the dial.

Then swap out for your unknown, and read the difference on the dial. As mentioned, the parts

*must* be at the same temperature, and the fixture must be isothermal over the course of the measurement.


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Reply to
jim rozen

Make a dial indicator holder so that you can simply compare the length of the two pieces. Depending on what you need, you could make it of non-metalic material to minimise temperature issues. It needs to be pretty stiff. Accuracy would then be about what the accuracy of your dial indicator is. Since you are measuring both pieces with the same tool at the same time, and since this is a purely comparative technique, you should be pretty accurate. I have one that I use to size flywheels bigger than my 12" dial caliper, and it is certainly repeatable to within a thou. Kind of the same idea as a dial bore gauge, but inside out. Mine cost $15 for the dial gauge and half an hour to cut and weld up the jig.

Use a firm joint caliper, if you want to go back to the old fashioned way of doing things. they can be good down to a few thou., maybe less, but depend to a great degree on operator skill (or lack thereof, in my case).

Make up big go/no go snap gauges. If you were doing a lot, this would be a good way to go.

All of these depend on having a sample to work, of course. Probably lots of other ideas out there, but all varioations on a theme.

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The e-mail address I have for you doesn't work any more. Would you send me one that works off line?

Thanks Fitch

Reply to
Fitch R. Williams

I would tell your customer that making something to "exact" specificiations is subject to a number of variables, such as temperature, and MEASURING TECHNIQUE. Therefore, your price is predicated on his coming to your shop and making the measurements as you aproach the finished length.

Reply to
Leo Lichtman

Yer fooling self ... yer 2' +/-.001 is only valid til the temp changes a degree or so. In fact, holding the bar in yer mitts will do it.

Reply to
Hoyt McKagen

snipped-for-privacy@hotmail.com (Ben) wrote in news:bc436b41.0312020944.2f06cde2 @posting.google.com:

Also accomplished via surface plate and height gauge.

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Except that doing it in pieces you'd end up with a possible error of the sum of the possible errors of each of the pieces. With the 8" dial calipers, good to

+- .001", you'd need four pieces - and would thus be +- .004". Probably better, but that is all you could guarantee.

Without gauge blocks or a large vernier gauge, cheapest way would probably be to buy a micrometer standard and compare to that. My 1985 Starrett price list shows a 24" standard for $34.45. Larger sizes available on special order.

John Martin

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A bit of exageration but mot too much. Ceof of linear expansion for yellow brass is 10.5E6/Fdegree. So you'd need 4 or so degrees to .001 on 26". ^ Hanbook of Chemistry and Physics, 69th ed.


Reply to
Ted Edwards

The chance that all the errors are in the same direction is extremely low. The usual method to calculate errors is the RMS average of the errors. That is you square all the errors and add them together and then take the square root. This assumes that all the errors are orthogonal. Or in this case the error you could reasonably expect is

+/- .002


Reply to
Dan Caster

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