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# Thread: Extending the range of calipers

1. ## Extending the range of calipers

Tony Foale has described a radius gauge...

based on the principle of measuring a chord and its related sagitta and then using these values to compute the radius of the measured curve.

The same principle can be applied to measure the diameter of circular objects whose diameter exceeds the capacity of your calipers. (Capacity here can refer to: calipers whose jaws do not open far enough to encompass the diameter to be measured or calipers whose jaws are too short to reach the diameter of the object.)

The first photo shows a too small caliper attempting to measure the diameter of an object. The caliper's jaws are too short to reach the diameter so it ends up measuring the length of some chord of the circle formed by the object. Note how the object is touching the beam of the caliper; it's pushed as far into the gape of the caliper as it will go.

This diagram will help to understand the mathematical notation...

The tips of the caliper jaws are at locations B and D. The line BD is the chord and that is what the calipers are measuring. The sagitta, EC (marked S in the diagram) is the distance from the chord to the circle's circumference. It should be obvious that that is the length of the caliper's jaws. Just to be clear, the following picture shows what you need to measure to find the length of your caliper's jaws...

(but use calipers not a scale; the picture is meant to show what to measure, not how to measure it.)

Now, if you know the length of the chord and the sagitta, you can compute the radius of the curve from a simple formula...

r = (c^2 + 4 * s^2) / (8 * s)

d = 2 * r = (c^2 + 4 * s^2) / (4 * s)

c = chord
s = sagitta
d = diameter

How well does it work ? Well, of course, that depends on your instruments and how carefully you measure.

For the bench block pictured, I made the following measurements...
c = 2.859
s = 1.159

and computed a diameter of 2.922. I then used my height gauge to measure the diameter (the jaws of my larger calipers are too short) and got a diameter of 2.924. YMMV.

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3. Marv,

Do you have any error analysis?

Rick

4. Originally Posted by rgsparber
Marv,

Do you have any error analysis?

Rick
Our mathmetician's mathematician, Marv don't need no stinkin' error analysis, lol.
But he provides next best thing "YMMV".

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that_other_guy (Apr 24, 2019)

7. Originally Posted by rgsparber
Marv,

Do you have any error analysis?

Rick
In the PDF from my post, that Marv mentioned, I wrote about that aspect. The same considerations apply in general to Marv's clever variation on that theme. The PDF is here RadiusGauge.pdf

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Scotsman Hosie (Apr 12, 2019)

9. Originally Posted by mklotz
Tony Foale has described a radius gauge...

based on the principle of measuring a chord and its related sagitta and then using these values to compute the radius of the measured curve.

The same principle can be applied to measure the diameter of circular objects whose diameter exceeds the capacity* of your calipers. (Capacity here can refer to: calipers whose jaws do not open far enough to encompass the diameter to be measured or calipers whose jaws are too short to reach the diameter of the object.)

The first photo shows a too small caliper attempting to measure the diameter of an object. The caliper's jaws are too short to reach the diameter so it ends up measuring the length of some chord of the circle formed by the object. Note how the object is touching the beam of the caliper; it's pushed as far into the gape of the caliper as it will go.

This diagram will help to understand the mathematical notation...

The tips of the caliper jaws are at locations B and D. The line BD is the chord and that is what the calipers are measuring. The sagitta, ED (marked S in the diagram) is the distance from the chord to the circle's circumference. It should be obvious that that is the length of the caliper's jaws. Just to be clear, the following picture shows what you need to measure to find the length of your caliper's jaws...

(but use calipers not a scale; the picture is meant to show what to measure, not how to measure it.)

Now, if you know the length of the chord and the sagitta, you can compute the radius of the curve from a simple formula...

r = (c^2 + 4 * s^2)(8 * s)

d = 2 * r = (c^2 + 4 * s^2)(4 * s)

c = chord
s = sagitta
d = diameter

How well does it work ? Well, of course, that depends on your instruments and how carefully you measure.

For the bench block pictured, I made the following measurements...
c = 2.859
s = 1.159

and computed a diameter of 2.922. I then used my height gauge to measure the diameter (the jaws of my larger calipers are too short) and got a diameter of 2.924. YMMV.
Sorry for being a bit thick but what does the ^ mean I've not seen this symbol used before

10. Originally Posted by joneb
Sorry for being a bit thick but what does the ^ mean I've not seen this symbol used before
It stands for "raised to the power". As examples...

x^2 means x-squared
x^3 means x-cubed
x^n means x raised to the n power

In handwritten math the exponent would just be written as superscript. Superscripts and subscripts aren't common on plain keyboards so many programming languages use the circumflex that is on most keyboards to indicate exponentiation.

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12. Originally Posted by rgsparber
Marv,

Do you have any error analysis?

Rick
Well, since we have the complete relationship of the variables expressed as an equation, constructing an error analysis is straightforward although it does require a bit of differential calculus.

We have from my previous post:

d = c^2/(4s) + s

Taking the derivative of d wrt c we have:

Dc = dd/dc = c/(2s)

and the derivative wrt s is:

Ds = dd/ds = 1 - c^2/(4s^2)

The error in d due to an error of magnitude Ec in c, which we'll label Ed, is given by:

Edc = Dc * Ec

Similarly, the error in d due to an error Es in s is given by:

Eds= Ds * Es

The total error in d is then given by the RSS (root-sum-square) of Edc and Eds...

Ed = sqrt (Edc^2 + Eds^2)

As my freshman calculus teacher used to say, it's left as an exercise for the student to fill in the numbers and obtain a final numerical value.

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Paul Jones (Apr 20, 2019)

14. Hi Marv,
That is a neat way to get the diameter of a cylinder that is larger than the capacity of your calipers. However in the above equation, the first term in brackets needs to be divided by the second term in brackets. The division sign between the two terms in the equation above is missing. Should be as follows:
d = 2*r = (c^2+4*s^2)/(4*s) or simply d = c^2/4*s +s
Thanks for the idea.
MrMetal

15. Originally Posted by MrMetal
Hi Marv,
That is a neat way to get the diameter of a cylinder that is larger than the capacity of your calipers. However in the above equation, the first term in brackets needs to be divided by the second term in brackets. The division sign between the two terms in the equation above is missing. Should be as follows:
d = 2*r = (c^2+4*s^2)/(4*s) or simply d = c^2/4*s +s
Thanks for the idea.
MrMetal
Thanks for that. I have no idea how that sign disappeared. I'll fix it immediately.

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Paul Jones (Apr 20, 2019)

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