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Mateyhv
member
Reged: 10/10/09
Posts: 57
Loc: 43°N
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Could anyone tell how could be calculated the cardbord masking circles for the lens so as to increase the f/ by 1/3 stop, 2/3, 1, etc for a given lens diameter? I suppose it has a direct relation with the surface of the lens as 1 stop allows half the light to enter the lens so half the surface of the lens has to be masked. Do someone have the equation?
Matey
Edited by Mateyhv (10/14/09 08:09 PM)
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EdZ
Professor EdZ
Reged: 02/15/02
Posts: 14732
Loc: Cumberland, R I , USA42N71.4W
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unless you know the F/# of the lens, you can't work in stops by fractions of a stop.
However, you can easily increase the total f# by 1/4 or 1/3 or 1/2 (1.25 or 1.33 or 1.5), meaning you can increase from f/4 to f/5 to f/5.33 or f/6
the inverse of the increase in f stop is the percent area of lens to leave exposed.
to increase f# by 1/4 or 1.25, then use lens of 1/1.25 = 0.8 or 80% the original diameter.
to increase f# by 1/3 or 1.33, then use lens of 1/1.33 = 0.75 or 75% the original diameter.
to increase f# by 1/2 or 1.50, then use lens of 1/1.50 = 0.67 or 67% the original diameter.
edz
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Teach a kid something today. The feeling you'll get is one of life's greatest rewards.
member#21
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Mateyhv
member
Reged: 10/10/09
Posts: 57
Loc: 43°N
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Hi edz, thanks for the answer, I edited my previous post because it was a bit unclear.
By the way when you say "to increase f# by 1/4 or 1.25, then use lens of 1/1.25 = 0.8 or 80% the original diameter." in fact you mean 80% the original area?
Not sure but I think that increasing 1 stop or 1 full f/ means reducing the light entering the lens by 1/2 and that is achieved by reducing the lens area by 1/2.
In a mask as the one showed by Philip Levine if the inner circle is cut at half the lens diameter it would yeld only 1/4 of the original area and not 1/2.
Using the formula of the area of a circle I got that 63mm lens has a total area of 3117 mm2. To increase f/ by 1 stop only 1558 mm2 are needed so to achieve that, the inner circle must have 45.4 mm diameter.
Please correct me if I am wrong.
Matey
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EdZ
Professor EdZ
Reged: 02/15/02
Posts: 14732
Loc: Cumberland, R I , USA42N71.4W
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Quote:
By the way when you say "to increase f# by 1/4 or 1.25, then use lens of 1/1.25 = 0.8 or 80% the original diameter." in fact you mean 80% the original area?
No. I meant diameter. f# is not calculated on area of lens
Quote:
Not sure but I think that increasing 1 stop or 1 full f/ means reducing the light entering the lens by 1/2 and that is achieved by reducing the lens area by 1/2.
That is an incorrect assumption. It means changing the ratio of F/D by 1.
Example: f/6 D = 80mm scope F= 480mm ,increase by 1f.
The current f# of this scope is found as 480/80 = f/6. Since we cannot change the focal length of the lens, then to increase this scope by 1f would mean to reduce the diameter of the aperture such that 480 divided by the new aperture = 7. So 480/7 = 68.6mm. Now we can see that the area has been reduced to 68.6 sqrd/80sqrd = 73.5% of it's original area. So area of the lens has been reduced by ONLY 26.5%.
Example2: f/4 D = 50mm binocular F= 200mm, increase by 1f.
The current f# of this scope is found as 200/50 = f/4. Since we cannot change the focal length of the lens, then to increase this scope by 1f would mean to reduce the diameter of the aperture such that 200 divided by the new aperture = 5. So 200/5 = 40mm. Now we can see that the area has been reduced to 40 sqrd/50sqrd = 64% of it's original area. So area of the lens has been reduced by 34%.
Example3: f/10 D= 125mm SCT F= 1250mm, increase by 1f.
The current f# of this scope is found as 1250/125 = f/10. Since we cannot change the focal length of the lens, then to increase this scope by 1f would mean to reduce the diameter of the aperture such that 1250 divided by the new aperture = 11. So 1250/11 = 113.6mm. Now we can see that the area has been reduced to 113.6 sqrd/125sqrd = 83% of it's original area. So area of the lens has been reduced by 17%.
As you can see from these examples, for a longer (slower) starting f#, changing the f# by 1 stop has less effect on the total area.
Quote:
In a mask as the one showed by Philip Levine if the inner circle is cut at half the lens diameter it would yeld only 1/4 of the original area and not 1/2.
true 1/2 the diameter would yeaild 1/4 the area, but 1/2 the diameter would DOUBLE the f# from (assumed) f/4 to f/8.
edz
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Teach a kid something today. The feeling you'll get is one of life's greatest rewards.
member#21
Edited by EdZ (10/15/09 06:18 AM)
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KennyJ
Reged: 04/27/03
Posts: 12904
Loc: Lancashire UK
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If I'm following these examples correctly , then put very simply :
If a 50mm binocular has a focal ratio of f/3 , then masking it down to 25mm would result in it being f/6 .
Or is a 70mm bino has a focal ratio of f/4 , then masking it down to 35mm would result in it being f/8 .
Would that be correct ?
Kenny
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EdZ
Professor EdZ
Reged: 02/15/02
Posts: 14732
Loc: Cumberland, R I , USA42N71.4W
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that is correct. Cutting aperture in half will doble the f#. makes sense.
F/D = Focal length / diameter = f#
edz
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Teach a kid something today. The feeling you'll get is one of life's greatest rewards.
member#21
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Mateyhv
member
Reged: 10/10/09
Posts: 57
Loc: 43°N
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Thanks Edz, I get it now. I see why I couldnt understand you at first, from my post it wasn't clear that I meant 0.1/3, 0.1/2 and 0.2/3 of 1 f/ stop as digicams do, so it was my fault.
So the problem for doing a precise masking device resides in that focal length or f/ value is not given for standart binoculars like mine Celestron 9x63.  Only can be done in fraction steps of the original unknown f/ value, as in your example (1.25, 1.33, 1.50...).
Can we go arround and for example calculate focal length by any other means (as FOV) so to get the correct f/ value? 
Matey
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Gordon Rayner
scholastic sledgehammer
Reged: 03/24/07
Posts: 967
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I recall noticing that the bottom of Crux is double, with a mounted WW II 10 x 80 in Oaxaca after the 1970 eclipse. I did not know its duplicity beforehand. I recall that it was tight, but do not recall any details of separation or non-separation, etc. I have not now researched the separation, so do not know if that was a normal observation or some kind of a feat. The components are bright. That sky area was fun to explore for the first time.There were several other eclipse travellers enjoying the southern sky for their first times. I wonder if the region near Suchixtepec, a small Indian village, is still dark and/or politically/narcotically safe for the tourist, particularly the lone tourist? I spoke little/no Spanish then, but there were European, Soviet, and North American astronomers nearby.
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EdZ
Professor EdZ
Reged: 02/15/02
Posts: 14732
Loc: Cumberland, R I , USA42N71.4W
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Quote:
I recall noticing that the bottom of Crux is double, with a mounted WW II 10 x 80 in Oaxaca after the 1970 eclipse. I did not know its duplicity beforehand. I recall that it was tight, but do not recall any details of separation or non-separation
Well the bottom of Crux, Alpha Cruxis, is a 4 arcsecond double. Being an accomplished double star observer myself, I would estimate at least a need for 30 to 40 power to see such a double. It does have a third component at a very wide separation of 90" which could very easily be seen at binocular powers.
edz
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Teach a kid something today. The feeling you'll get is one of life's greatest rewards.
member#21
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