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MKV
Carpal Tunnel
Reged: 01/20/11
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Re: DIY Spherometer... Wood?
[Re: Pinbout]
#5251292 - 06/01/12 11:28 PM
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I got this from Mr. Parker,
...the formulas for spherometer with ball feet are supposed to be:
S=(R-d/2)-((R-d/2)^2-r^2)^0.5 for concave surfaces
S=(R+d/2)-((R+d/2)^2-r^2)^0.5 for convex surfaces
Where S = sagitta, R = ROC, d = diameter of ball foot, r = radius of spherometer base circle.
Danny, Mr. Parker's equation is a formula the sagitta; Footbag's equation is the formula for the radius of curvature - apples and oranges.
Quote:
...it is important to know the diameter (or radius) of your spherometer foot circle pretty precisely. Precise spherometers are more important for lens making where the ROC should be within 1/20% of the design value. Also for best results, the spherometer foot circle should be as large as you can get....
A 1/20% is a very tight tolerance in good aprt because of the internal errors of measuring equipment.
For a radius of curvature of 100 inches this means it could not vary by more than +/-0.025 inches (a total range of 0.05 or 1.27 mm in length).
Let's say you have a 100 mm diameter spherometer, and you know the distance of its feet to within 0.01 mm (0.0004 in). And let's assume you have a micrometer head or a dial indicator reading sagitta to a precision of +/- 0.001 mm (0.00004 in)). Such a spherometer would still have an internal error at least 10 times greater then the the 0.05% range for the 100 inch roc!
Mladen
Edited by MKV (06/02/12 05:17 AM)
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Footbag
Carpal Tunnel
Reged: 04/13/09
Loc: Scranton, PA
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Re: DIY Spherometer... Wood?
[Re: MKV]
#5251691 - 06/02/12 08:31 AM
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I do have ball feet on my spherometer and they are 1/4" diameter. I also used a caliper to determine the distance between the legs. My caliper is plastic and non-digital and cost $1.99 at Harbor Freight to give you an idea of precision.
I have three formulas to get me from the raw measurements to the RoC. I used Gordon Waite's formulas from his video here.
First, I take the distance between each of the three ball feet and subtract the diameter of the actual feet. (This is what I was referring to above when I said I already removed the diameter of the ball feet) This gives me A, B and C. Then I use the following forumla...
K=1/2(A+B+C)
This gives me K, which Gordon describes as a constant used to determine the Radius of the spherometer in the next formula.
R= ABC/4^(1/2)K(K-A)(K-B)(K-C)
Then I use my sagitta reading(S), the radius of the spherometer(R) and the following formula to determine the radius of curvature of the mirror.
Rm=Rs^2-S^2/2S + d/2
Obviously, once I get adequately precise measurements for the distance between the ball feet, I can calculate the radius of the spherometer and won't have to redo the first two equations.
Also, I did realize that by moving the ball feet out from the center, I would get a more accurate reading. But, the socket head bolts I'm using are as close to the edge of the groove of the pulley wheel as possible. My only other option would be to lap the bottom of the pulley wheel (to level it)and use the outside edge to rest on the mirror. If I did that, then determining the radius of the mirror would be easy (5")and I could skip the first two formulas. I think the difficulty is determining the radius of the ball feet which may or may not be evenly spaced.
Thanks again.
Edited by Footbag (06/02/12 08:32 AM)
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Pinbout
Post Laureate
Reged: 02/22/10
Loc: Montclair
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Re: DIY Spherometer... Wood?
[Re: Footbag]
#5252506 - 06/02/12 07:46 PM
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This gives me K, which Gordon describes as a constant used to determine the Radius of the spherometer in the next formula.
the constant K is really a formula for a semi-perimeter of the triangle formed by the feet.
ref found here...
http://en.wikipedia.org/wiki/Circumscribed_circle
they use the variable "s".
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Footbag
Carpal Tunnel
Reged: 04/13/09
Loc: Scranton, PA
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Re: DIY Spherometer... Wood?
[Re: Pinbout]
#5252521 - 06/02/12 07:57 PM
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Quote:
This gives me K, which Gordon describes as a constant used to determine the Radius of the spherometer in the next formula.
the constant K is really a formula for a semi-perimeter of the triangle formed by the feet.
ref found here...
http://en.wikipedia.org/wiki/Circumscribed_circle
they use the variable "s".
Yup. That's how I understand it. So being that I removed them before plugging ABC in to the first formula, I was thinking that I may not have to use them in the last formula.
I understand what each formula is doing, but I'm not quite sure I understand the + or - d/2 where d is the diameter of the balls.
Do the ball bearings cause the spherometer to rest on the glass in a different way then pointed feet do? And in that case, do larger ball bearings rest in a different place then small ones would? Isn't that what the formula is implying?
EDIT: Got it! Yes they do...
Edited by Footbag (06/02/12 08:14 PM)
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MKV
Carpal Tunnel
Reged: 01/20/11
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Re: DIY Spherometer... Wood?
[Re: Footbag]
#5252832 - 06/03/12 12:33 AM
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I do have ball feet on my spherometer and they are 1/4" diameter. I also used a caliper to determine the distance between the legs. My caliper is plastic and non-digital and cost $1.99 at Harbor Freight to give you an idea of precision...
How much is that (it should be listed)? Also, what is the average diameter of the legs, and what precision is your dial indicator?
Mladen
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John Carruthers
Skiprat
Reged: 02/02/07
Loc: Kent, UK
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Re: DIY Spherometer... Wood?
[Re: Footbag]
#5252948 - 06/03/12 06:43 AM
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larger/smaller = yes you are correct. I use ball pen tips (very hard)at 2" radius (easy maths). Sharp points can be more accurate but they wear and can sit in pits during coarse grinding.
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