From: "Rich Ball" To: "scottythefiddler" Reply-To: "Rich Ball" This is a multi-part message in MIME format. ------=_NextPart_000_009B_01C32D3C.D5DA3140 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Scotty the fiddler -- (et al.), Hey, I thank you very much. I looked at the mirror and I think I'm in = for some work. There seems to be too much parabola. My figures are as = follows ( as deviations from your figures): =20 actual: zone 1 1.264 * 1.264 / 99.5 =3D 0.016" .016" =20 zone 2 2.192 * 2.192 / 99.5 =3D 0.048" .065" zone 3 2.828 * 2.828 / 99.5 =3D 0.080" .125" zone 4 3.348 * 3.348 / 99.5 =3D 0.113" .260" zone 5 3.780 * 3.780 / 99.5 =3D 0.144 .425" I found zone 1 and set my scale at .016. Then I found the other zones = at the inch-marks indicated. They came up with a satisfying lack of = ambiguity (perhaps because they are too widely spaced). I did the test = a few times and came up with small variations. =20 If I read theses numbers correctly I would guess that I must work = carefully around the perimeter of the mirror, especially, in order to = shallow-up the paraboliod figure. No? Full sized tool? Tool on top or = mirror on top? I'll have to read up on this one. Curiously the current figure came up without trying to deepen the figure = during polishing. I made no effort to deepen the sphere. I was almost = lucky. --Rich Ball (time out to think) You know - I just realized there is something I don't know and have = taken for granted. I presumed the numbers that I hope to find on my = Foucault tester were like mile markers on a highway, the first mile = comes up at .016 and the second at mile .048 etc. But it just occured = to me that these numbers could be distances between zones. Is this = clear? That is a huge difference. Because (DUH!) if the numbers are = not additive distances my figure is much closer to my goal. It would be.......................................... actual zone 1 1.264 * 1.264 / 99.5 =3D 0.016" .016" zone 2 2.192 * 2.192 / 99.5 =3D 0.048" .032" zone 3 2.828 * 2.828 / 99.5 =3D 0.080" .060" zone 4 3.348 * 3.348 / 99.5 =3D 0.113" .135" zone 5 3.780 * 3.780 / 99.5 =3D 0.144 .190" I think I may be an idiot!! I hope so. Please confirm. I will be a = happy idiot. =20 --Rich Ball =20 ----- Original Message -----=20 From: scottythefiddler=20 To: Rich Ball=20 Sent: Saturday, June 07, 2003 8:33 PM Subject: Re: ATM figuring an 8" f6 mirror I assume that you used r^2/2R to calculate your longitudinal = abberation (deviation from C of C)... However, for a fixed source, the = longitudinal abberation is twice that.....r^2/R (This has = sometimes been referred to as the 'fog of two's', i.e. common error , = don't ask me how I know...lol) zone 1 1.264 * 1.264 / 99.5 =3D 0.016" zone 2 2.192 * 2.192 / 99.5 =3D 0.048" zone 3 2.828 * 2.828 / 99.5 =3D 0.080" zone 4 3.348 * 3.348 / 99.5 =3D 0.113" zone 5 3.780 * 3.780 / 99.5 =3D 0.144" Best of luck with your scope. I am sure that it will be a good one. = Tell us more about it. Scott Donaldson ----- Original Message -----=20 From: Rich Ball=20 To: atm{at}shore.net=20 Sent: Saturday, June 07, 2003 5:57 PM Subject: ATM figuring an 8" f6 mirror Help! I am testing the figure on my 8" mirror which has a radius of = curvature of 99.5 inches. I am using my homemade Foucault tester which = has a fixed light source. If there is a soul out there who could critique my math I'd be much = beholden. I am working with 5 zones, perhaps overkill. And the spacing I = calculated for a Couder mask (A) and the ideal deviations between zones = (B) are as follows: Zone ctr deviation=20 radius fr C of C=20 (A) (B)=20 Zone 1 1.264 0.008=20 Zone 2 2.192 0.024=20 Zone 3 2.828 0.040=20 Zone 4 3.348 0.056=20 Zone 5 3.780 0.072=20 Does this look right? I seem to have some paraboloid figure to = begin with. But this stuff is very hard for this neophyte. Thank you. Rich Ball PS. Thank you to the kind soul who told me about David Harbour's = fine article on Foucault testing. I recommend it to any neophyte at = www.atm.org/contrib/Harbour/Foucault.html ------=_NextPart_000_009B_01C32D3C.D5DA3140 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Scotty the fiddler -- (et = al.), Hey, I thank you very much. I looked at = the=20 mirror and I think I'm in for some work. There seems to be too = much=20 parabola. My figures are as follows ( as deviations from your=20 figures): &nb= sp; &nbs= p; = ; = &= nbsp; =20 =20 = =20 = =20 = =20 actual: zone 1 1.264 * 1.264 = / 99.5 =3D=20 0.016" = .016" =20 zone 2 2.192 * 2.192 / = 99.5 =3D=20 0.048" .065" zone 3 2.828 * 2.828 = / 99.5 =3D=20 0.080" = .125" zone 4 3.348 * = 3.348 / 99.5=20 =3D 0.113" = .260" zone 5 3.780 * 3.780 = / 99.5 =3D=20 0.144 .4= 25" I found zone 1 and set my scale at .016. Then I found the = other zones=20 at the inch-marks indicated. They came up with a satisfying lack = of=20 ambiguity (perhaps because they are too widely spaced). I did the = test a=20 few times and came up with small variations. If I read theses numbers correctly I would guess that I must work = carefully=20 around the perimeter of the mirror, especially, in order = to shallow-up=20 the paraboliod figure. No? Full sized tool? Tool on top or = mirror on=20 top? I'll have to read up on this one. Curiously the current figure came up without trying to deepen the = figure=20 during polishing. I made no effort to deepen the sphere. I = was=20 almost lucky. --Rich Ball (time out to think) You know - I just realized there is something I don't know and have = taken=20 for granted. I presumed the numbers that I hope to find on my = Foucault tester were like mile markers on a highway, the first mile = comes=20 up at .016 and the second at mile .048 etc. But it just occured to = me that=20 these numbers could be distances between zones. Is this = clear? That=20 is a huge difference. Because (DUH!) if the numbers are not = additive=20 distances my figure is much closer to my goal. It would be..........................................=20 actual zone 1 1.264 * 1.264 = / 99.5 =3D=20 0.016" .016" zone 2 2.192 * 2.192 / = 99.5 =3D=20 0.048" .032" zone 3 2.828 * 2.828 = / 99.5 =3D=20 0.080" .060" zone 4 3.348 * = 3.348 / 99.5=20 =3D 0.113" .135" zone 5 3.780 * 3.780 = / 99.5 =3D=20 0.144 .190" I think I may be an idiot!! I = hope so. =20 Please confirm. I will be a happy idiot. --Rich Ball =

----- Original Message ----- From:=20 scottythefiddler">mailto:scottythefiddler{at}cogeco.ca">scottythefiddler To: Rich">mailto:ShooterBall{at}hotmail.com">Rich Ball Sent: Saturday, June 07, 2003 = 8:33=20 PM
Subject: Re: ATM figuring an 8" = f6=20 mirror I assume that you used r^2/2R to = calculate your=20 longitudinal abberation (deviation from C of C)... However, for = a fixed=20 source, the longitudinal abberation is twice=20 that.....r^2/R (This has sometimes = been=20 referred to as the 'fog of two's', i.e. common error , don't ask = me how=20 I know...lol) zone 1 1.264 * = 1.264 / 99.5 =3D=20 0.016" zone 2 2.192 * 2.192 / = 99.5 =3D=20 0.048" zone 3 2.828 * = 2.828 / 99.5 =3D=20 0.080" zone 4 3.348 * = 3.348 /=20 99.5 =3D 0.113" zone 5 3.780 * = 3.780 / 99.5 =3D=20 0.144" Best of luck with your = scope. I am=20 sure that it will be a good one. Tell us more about = it. Scott Donaldson
----- Original Message ----- From:=20 Rich=20">mailto:ShooterBall{at}hotmail.com">Rich=20 Ball To: atm{at}shore.net">mailto:atm{at}shore.net">atm{at}shore.net
Sent: Saturday, June 07, 2003 = 5:57=20 PM
Subject: ATM figuring an 8" = f6=20 mirror Help! I am testing the figure on my 8" mirror = which has a=20 radius of curvature of 99.5 inches. I am using my homemade = Foucault=20 tester which has a fixed light source. If there is a soul out there who could = critique my=20 math I'd be much beholden. I am working with 5 zones, perhaps = overkill. =20 And the spacing I calculated for a Couder mask (A) and the = ideal=20 deviations between zones (B) are as follows:

Zone=20 ctr deviation radius fr C of C (A) (B) Zone 1 =20 1.264 =20 0.008 Zone 2 =20 2.192 =20 0.024 Zone 3 2.828 =20 0.040 Zone 4 =20 3.348 =20 0.056 Zone 5 = 3.780 =20 0.072 Does this look right? I seem to = have some=20 paraboloid figure to begin with. But this stuff is very hard = for this=20 neophyte. Thank you. Rich Ball PS. Thank you to the kind soul who = told me=20 about David Harbour's fine article on Foucault testing. I = recommend it=20 to any neophyte at www.atm.org/con=">http://www.atm.org/contrib/Harbour/Foucault.html">www.atm.org/con= trib/Harbour/Foucault.html ------=_NextPart_000_009B_01C32D3C.D5DA3140-- --- BBBS/NT v4.01 Flag-4 * Origin: Email Gate (1:379/100)
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