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Challenge Question - Least Count of a Vernier Caliper Consider a Vernier Caliper with - m M.S.D = v V.S.D; where m,v are co-prime (v can also be less than m) Find its least count. Note that many websites (including Wikipedia) gives - L.C = 1 M.S.D - 1 V.S.D http://en.wikipedia.org/wiki/Least_count but I beg to differ
Just to be precise - The smallest value that can be measured by the measuring instrument is called its least count. PS: As already stated, this is a challenge question. Because I already know the answer (or atleast have such an impression) "providing answers" with/without solution may not be against the CoC (Still I would prefer the moderators to clarify in this regard).
the main scale division will have some least count.. now depending on the least count of main scale, the vernier scale is set up.. for example if the main scale is a CM scale with 10 divisions in between (like a normal scale), then its vernier may have 11 divisions squeezed up in that same space, (giving 10 vernier scale readings 0 to 10). so the main scale has least count of 1/10 = 0.1 cm, and vernier itself gives 10 more divisions thus total least count is 0.1/10 = 0.01 cm usually moving microscopes have main scale having 20 divisions between a cm.. so giving it LC = 0.05cm, and vernier ends up having 51 divisions squeezed up in the place of 2.5 cm decreasing LC by a factor of 50.. thus total L.C. = 0.05/50 = .001 cm
LC of main scale = 1 M.S.D LC of vernier scale = 1 V.S.D (taken separately) "...if the main scale is a CM scale with 10 divisions in between (like a normal scale), then its vernier may have 11 divisions squeezed up in that same space..." In my notation, 10 M.S.D = 11 V.S.D "..microscopes have main scale having 20 divisions between a cm.. so giving it LC = 0.05cm, and vernier ends up having 51 divisions squeezed up in the place of 2.5 cm ..." In my notation, 50 M.S.D = 51 V.S.D ^^ The above was to clarify my (shorthand) notation. (As Wikipedia uses the same, I considered them to be standard notations.) @Mashy I don't know much about microscopes, but I think they use the concept of a screw gauge. In case of a Vernier Caliper, I can prove (by elementary mathematics) that LC is less than what you calculated. For the more common cases (and I think you were trying to refer to these) 9 M.S.D = 10 V.S.D ; LC = 0.1 M.S.D 49 M.S.D = 50 V.S.D ; LC = 0.02 M.S.D
I won't address any central point -- but this is fascinating. I'll be watching this thread.
Bump in 645:52... really ?? I think I'll answer it myself before the next bump :'(
Answer:\[L.C.=\frac{ 1 }{ v } M.S.D.\] Proof (Let m<v) - Consider a point on the main scale at 'a' units - |dw:1390023259606:dw| Note: [.] implies greatest integer function {.} implies fractional parts As the diagram shows, the distance between the corresponding divisions of Main Scale and Vernier Scale (note: the expression in the diagram is not properly written; refer to the one below) - \[=a-\left[ \frac{ av }{ m } \right]*\frac{ m }{ v } M.S.D.\] \[=\frac{ m }{ v }\left\{ \frac{ av }{ m } \right\} M.S.D\] As the minimum value of the function \[f(x)=\left\{ \frac{ ax }{b } \right\} \] where a,b are co-prime and x∈N \[=\frac{ 1 }{ b} (why?)\] \[L.C.=\frac{ 1 }{ v } M.S.D.\] By symmetry, the proof can also be extended to m>v.
See the next question - http://openstudy.com/study#/updates/52da2561e4b0b71c4f8dcf75
^^ It is much easier than this one. :)