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Re: [ATM] parabolic versus spherical volume of glass
Don,
That is interesting stuff. I think you have more fun
than I do...
Vlad
----- Original Message -----
From: Donald Good<mailto:donald.good@comcast.net>
To: 'vladimir sacek'<mailto:v959@msn.com> ; atm@atmlist.net<mailto:atm@atmlist.net>
Sent: Friday, January 07, 2005 12:14 AM
Subject: RE: [ATM] parabolic versus spherical volume of glass
Revised FYI
FYI, for the example given in the spreadsheet (D=16.5, F/5), the
values of fl diff for the parabolas described by Vlad are:
parabola fl diff vol diff min sag Vlad's sag values for (D=16.5,
F/5)
#1 0 .00920 0%R D/1024F^3=.000129 @r=R
#2 -.0515 .00230 70%R D/4000F^3=.000033 @r=R
#3 -.0605 .00249 75%R D/6000F^3=.000022 @r=R
#4 -.104 .00943 100%R D/1000F^3=.000132 @r=0
#3 was changed from D/13,000F^3=.000010 @r=R to D/6000F^3=.000022 @r=R.
The vol diff, which is the minimum volume of glass removed for each
parabola, was reformulated in the spreadsheet. From Vlad's formula
V=Vp-Vs+x*Pi*d^2, the term x*Pi*d^2 is merely the volume of a cylinder of
radius (=semi-diameter) d and height x. I found that this term was hidden
in my calculation of the glass volume (J28, sum sag vol). The calculation
for the elements in J was for a cylindrical shell of thickness R/20, radius
R, and height h=max sag diff-sag diff at r with some refinements at r=0
(cylinder) and r=R (half shell). The volume was the sum of these shells.
This was just an approximation which becomes the true volume as N, the
number of shells (and points along r), approaches infinity and the thickness
R/N approaches 0 (the integral in calculus). Well the max sag difference is
a sufficiently close approximation of Vlad's x. Here it is found by
inspection (MAX in J3) of 21 points on r (increasing accuracy as the number
of plot points increases). So G4=Vp-Vs, J4=x*Pi*d^2 and G5=V. This
completely eliminates the "shell game" in J. The revised spreadsheet is at:
http://www.atmlist.net/contrib/donald-dot-good-at-comcast-dot-net/index.htm<http://www.atmlist.net/contrib/donald-dot-good-at-comcast-dot-net/index.htm>
The graph is not the shape of the mirror. It is merely the difference
between the sag (sagitta) of a spherical section and the sag of a
corresponding paraboloid section of the same sectional diameter at 21 points
along the sectional radius. This locks the 0 reference at the edge (r=R)
for both sphere and paraboloid. Where the graph is positive, the paraboloid
sag is less than the spherical sag and the paraboloid surface would be in
front of the spherical surface at that point. The opposite is true where
the graph is negative. But you can't add glass where the graph is positive
when polishing. Fortunately you don't have to. That is the purpose of the
term x*Pi*d^2. Mathematically, it is like "adding" a layer of glass to the
front of the sphere. Physically, it is just changing the reference point of
the parabola relative to the sphere along the optical axis by x.
Clear skies,
Don
> -----Original Message-----
> From: atm-bounces@atmlist.net<mailto:atm-bounces@atmlist.net>
> [mailto:atm-bounces@atmlist.net] On Behalf Of vladimir sacek
> Sent: Friday, January 07, 2005 2:20 PM
> To: atm@atmlist.net<mailto:atm@atmlist.net>
> Subject: Re: [ATM] parabolic versus spherical volume of glass
>
> Assuming that "differential volume" stands for the minimum
> volume of glass needed to be removed, we do have some discrepancies.
> I did my figures more precisely (I've being rounding a bit
> too much previously), and came up with these volumes:
> #1 0.0091 ci
> #2 0.0027
> #3 0.0065
> #4 0.0093
>
> One important correction is for parabola #3, which needs to
> have taken about 2/3 as much from the edge as #2 (not 1/3, as
> I posted previously), and proportion of digging in the center
> vs. edge is about 3 to 1 (not 6:1). Consequently, the depth
> of glass taken from the edge is ~D/6000F^3. Also, needed
> total volume of glass to remove is more than doubled vs. #2.
>
> The volume formula I use is V=Vp-Vs+x*Pi*d^2, where Vp is
> volume of the parabola, Vs volume of the sphere, "x"
> edge-to-edge separation between the sphere and parabola when
> they are touching at the given zonal hight, and "d" the
> aperture semi-diameter. The sphere and parabola volume
> formulas are identical to those you use, I believe, and
> x=(1/2Rs-1/2Rp)(1-z^2)D^2 + [(1-z^4)d^4)]/8Rs^3 with Rs and
> Rp being the respective curvature radii of sphere and
> parabolas, and "z" given zonal hight for which both sphere
> and parabola have a common focus.
>
> The bottom line I end up with is that the least amount of
> glass removed is for parabola #2, approximately D^3/13,000F^3.
> #3 requires more than twice, and #1 and #4 more than three
> times as much glass removed.
>
> Vlad
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