Re: [pinhole-discussion] f stops
- Original Message - From: Andy Schmitt aschm...@warwick.net Gill you have WAY too much time on your hands.. but a great explaination..I'll save this one thanks I am sorry Andy, was off, back was acting up, weather lousy, no reason to go out. From now on I will only reply when I don't have WAAAY too much time in my hands, so answers are as un-prolific as possible. Guillermo
RE: [pinhole-discussion] f stops
Gill you have WAY too much time on your hands.. but a great explaination..I'll save this one thanks andy -Original Message- From: pinhole-discussion-admin@p at ??? [mailto:pinhole-discussion-admin@p at ???]On Behalf Of Guillermo Sent: Wednesday, October 31, 2001 10:19 AM To: pinhole-discussion@p at ??? Subject: Re: [pinhole-discussion] f stops - Original Message - From: ragowaring ragowar...@btinternet.com They follow a geometric progression that make the f/stops increase by a factor of square root of 2 Wow!, is it really that simple. Of course, it all makes sense now and it will be very helpful. Thankyou for your replies. If it is not too much trouble a complete explanation would be very welcome Guillermo. Although John Yeo already gave you an explanation, risking being redundant, I will give you another one: In general terms, f/stop is just a ratio that tell us how many times the diameter of the aperture the focal length of your lens is. i.e.. a 50mm lens with an aperture diaphragm opening of 25mm in diameter would have an f/stop of f/2 (50/25 = 2). Another example: a 90mm focal length pinhole camera with a 0.35mm pinhole would have an f/stop = f/114 ( 90 / 0.35 = 114 ) f/stop = focal length / aperture diameter Both Focal length and Diameter must be given in the same units of measure. The amount of light that any f/stop let through is double the one the immediate closed down full f/stop let through. For instance, f/8 let through double the amount of light than f/11 let through and that means the area enclosed by the round aperture opening at f/8 is twice the one for f/11. The area enclosed by a circle is proportional to the square of its diameter (Area = 0.7854 * D^2), therefore, to double the area inside a circle (allowing double the light, meaning opening up 1 f/stop), we need to increase the diameter of the aperture by just square root of 2 = 1.414 We now know the diameter of the aperture from one stop to the next, increase by a factor of 1.414, we also know f/stop = focal length / diameter, therefore the f/stop numbers will increase also by a factor of 1.414 Starting with f/1 (which BTW is neither the theoretical nor practical maximum aperture), to find out the next full stop, just multiply the preceding one by 1.414 and approximate the result as required, like this: 1 = f/1 1 x 1.414 = f/1.4 1.4 x 1.414 = f/2 2 x 1.414 = f/2.8 2.8 x 1.414 = f/4 4 x 1.414 = f/5.6 5.6 x 1.414 = f/8 and so on. As you can notice, the numerical value of the f/stop doubles every other full f/stop, so after you find the first 2 (f/1 and f/1.4) you no longer need to multiply but 1.414 just double the f/stop 2 stops behind. ie. the next stop after f/8 would be equal to double f/5.6 = f/11 (5.6 x 2 = 11), the next stop after f/11 would be double f/8 = f/16 and so on. As an added information, I'd like to mention how to find out intermediate f/stops. When we divide the focal length by the pinhole diameter, most likely than not we get an f/stop number that is not a full f/stop. Some paragraphs above I mentioned the following example: a 90mm focal length pinhole camera with a 0.35mm pinhole would have an f/stop = f/114 ( 90 / 0.35 = 114 ), it is clear that we need to approximate that f/114 to either a full stop, 1/2 stop, 1/3 stop or whatever you like. I usually approximate to the next 1/3 or 1/2 stop, the question becomes: how to know which of these f/114 is closer to? Get your slide ruler or your scientific calculator (I use CALC98 http://www.calculator.org/) 'cause we will get some logarithms. To find where f/114 falls with respect to the closest larger full f/stop (larger means smaller numerical value, remember), we just divide f/114 by that larger full f/stop (f/90 in our case), then get the common logarithm (LOG in most calculators, as suppose to Ln) of the answer and finally divide that by 0.15 Let see: 114 / 90 = 1.26 LOG (1.26) = 0.10266 0.10266 / 0.15 = 0.684 The 0.684 means f/114 is 0.684 stops smaller than f/90 , knowing this helps me to know that f/114 is just a bit larger (numerically) than f/90 2/3 (2/3 = 0.666) and also f/114 is just smaller than f/90 3/4 stops (3/4 = 0.750). Knowing all this may be an overkill for most of us, but it doesn't hurt knowing it, anyway. I better stop here. Guillermo ___ Pinhole-Discussion mailing list Pinhole-Discussion@p at ??? unsubscribe or change your account at http://www.???/discussion/
Re: [pinhole-discussion] f stops
on 31/10/01 3:18 pm, Guillermo at pen...@home.com wrote: - Original Message - From: ragowaring ragowar...@btinternet.com They follow a geometric progression that make the f/stops increase by a factor of square root of 2 Wow!, is it really that simple. Of course, it all makes sense now and it will be very helpful. Thankyou for your replies. If it is not too much trouble a complete explanation would be very welcome Guillermo. Although John Yeo already gave you an explanation, risking being redundant, I will give you another one: In general terms, f/stop is just a ratio that tell us how many times the diameter of the aperture the focal length of your lens is. i.e.. a 50mm lens with an aperture diaphragm opening of 25mm in diameter would have an f/stop of f/2 (50/25 = 2). Another example: a 90mm focal length pinhole camera with a 0.35mm pinhole would have an f/stop = f/114 ( 90 / 0.35 = 114 ) f/stop = focal length / aperture diameter Both Focal length and Diameter must be given in the same units of measure. The amount of light that any f/stop let through is double the one the immediate closed down full f/stop let through. For instance, f/8 let through double the amount of light than f/11 let through and that means the area enclosed by the round aperture opening at f/8 is twice the one for f/11. The area enclosed by a circle is proportional to the square of its diameter (Area = 0.7854 * D^2), therefore, to double the area inside a circle (allowing double the light, meaning opening up 1 f/stop), we need to increase the diameter of the aperture by just square root of 2 = 1.414 We now know the diameter of the aperture from one stop to the next, increase by a factor of 1.414, we also know f/stop = focal length / diameter, therefore the f/stop numbers will increase also by a factor of 1.414 Starting with f/1 (which BTW is neither the theoretical nor practical maximum aperture), to find out the next full stop, just multiply the preceding one by 1.414 and approximate the result as required, like this: 1 = f/1 1 x 1.414 = f/1.4 1.4 x 1.414 = f/2 2 x 1.414 = f/2.8 2.8 x 1.414 = f/4 4 x 1.414 = f/5.6 5.6 x 1.414 = f/8 and so on. As you can notice, the numerical value of the f/stop doubles every other full f/stop, so after you find the first 2 (f/1 and f/1.4) you no longer need to multiply but 1.414 just double the f/stop 2 stops behind. ie. the next stop after f/8 would be equal to double f/5.6 = f/11 (5.6 x 2 = 11), the next stop after f/11 would be double f/8 = f/16 and so on. As an added information, I'd like to mention how to find out intermediate f/stops. When we divide the focal length by the pinhole diameter, most likely than not we get an f/stop number that is not a full f/stop. Some paragraphs above I mentioned the following example: a 90mm focal length pinhole camera with a 0.35mm pinhole would have an f/stop = f/114 ( 90 / 0.35 = 114 ), it is clear that we need to approximate that f/114 to either a full stop, 1/2 stop, 1/3 stop or whatever you like. I usually approximate to the next 1/3 or 1/2 stop, the question becomes: how to know which of these f/114 is closer to? Get your slide ruler or your scientific calculator (I use CALC98 http://www.calculator.org/) 'cause we will get some logarithms. To find where f/114 falls with respect to the closest larger full f/stop (larger means smaller numerical value, remember), we just divide f/114 by that larger full f/stop (f/90 in our case), then get the common logarithm (LOG in most calculators, as suppose to Ln) of the answer and finally divide that by 0.15 Let see: 114 / 90 = 1.26 LOG (1.26) = 0.10266 0.10266 / 0.15 = 0.684 The 0.684 means f/114 is 0.684 stops smaller than f/90 , knowing this helps me to know that f/114 is just a bit larger (numerically) than f/90 2/3 (2/3 = 0.666) and also f/114 is just smaller than f/90 3/4 stops (3/4 = 0.750). Knowing all this may be an overkill for most of us, but it doesn't hurt knowing it, anyway. I better stop here. Guillermo ___ Pinhole-Discussion mailing list Pinhole-Discussion@p at ??? unsubscribe or change your account at http://www.???/discussion/ Thankyou John Yeo and Guillermo for your exhaustive explanations. I new some of it but it seems now I will know all of it (or have I said the wrong thing). Never again shall I want for knowing. Thankyou again. All the best Alexis
