Love this series! It has helped understand exactly what is going on and what I should think about for my build. One thing I didn't find covered in these videos is optimal pinhole size. It seems like most people thing the smaller the better, but I wonder if that is true. Can you shed any light on this? Assuming a perfectly round shape and relatively thin material, how does a 1mm vs .5mm vs .25mm stack up? Is sharpness perceptibly better? Or will images still be soft? Also, do you have a website or other social media accounts?
Thanks! I have a video on optimal pinhole size here: ruclips.net/video/n5W3qztO4os/видео.html (it's a bit long, but it goes through all the math). If you want to see the impact of size on sharpness check out this video: ruclips.net/video/VPzAAzEb1wE/видео.html In short, there is an optimal value that you can calculate, but error on the side of smaller since the impact of diffraction isn't as bad as having a pinhole that is too large. my website is at CyrusArthur.com Let me know if you have any other questions
Would curving the film plane eliminate or reduce vignetting, since the distance from the pinhole to the focal 'plane' would be similar at all points, (assuming an equidistant curve)? Thanks for the video.
Great question. Yes, if you match the curvature to the angle of projection from the pinhole then vignetting is greatly reduced. To eliminate it completely you would need a spherical film plane. However, the image will no longer be rectilinear and will have some level of distortion. Film cameras such as the widelux and the noblex have curved film planes and produce an almost fisheye effect. To reduce vignetting on a flat surface use a longer focal length (or pinhole distance) so that the difference in distance to the center and edges of the image is less. Check out my more recent pinhole videos on my 8x10 superwide camera to see the impact of vignetting. It is quite pronounced in my sample images.
I’m not sure what I’m doing wrong. I have a pinhole diameter of .3 mm into thickness of .1 mm. When I use your formula I get a theta of 1.25 ? I can’t imagine my viewing angle is 2.5°?
Looks like your calculator is giving an answer in radians since tan^-1(.3/.1) = 1.249 radians so the total angle is ~2.5 radians. Converting 2.5 radians into degrees is about 143 degree angle of view. Hope that helps!
I'm doing these math operations to know what's the angle of view that will be proyect on a wall of my bath room(yes, i want to use the window of my bath room as the plane of the pinhole...jajaja) so i have these parameters f=1100mm d=1.2mm t=0.014mm(aluminium foil) And all is equal to Q=7,170.632701 What kind of number is that i mean what angle is that? Thanks i really apreciate your lesson
You need to take the inverse tangent of d/t so when do you the calculation it equals 1.5591 radians or 89 degrees. Multiply that by 2 to get 178 degrees which is basically an unlimited angle of coverage. You will have an f/stop of f/916 which will make the image very dim so you would have to let your eyes adjust for a long time.
I'm not sure I understand your question. I did have both d and t since I bought a pinhole of known diameter and known thickness. The expression you wrote in your comment simplifies to the one I used tan^-1(0.3/0.127) since both the numerator and denominator are divided by 2 it cancels out.
Concepts well explained 👏👏👏👏
After find out the minimum focal length, eg 50mm, so I can increase the focal length, eg 60mm, when I build the camera with same coverage??
If you increase the focal length from the minimum focal length you will get even more coverage
Love this series! It has helped understand exactly what is going on and what I should think about for my build. One thing I didn't find covered in these videos is optimal pinhole size. It seems like most people thing the smaller the better, but I wonder if that is true. Can you shed any light on this? Assuming a perfectly round shape and relatively thin material, how does a 1mm vs .5mm vs .25mm stack up? Is sharpness perceptibly better? Or will images still be soft?
Also, do you have a website or other social media accounts?
Thanks! I have a video on optimal pinhole size here: ruclips.net/video/n5W3qztO4os/видео.html (it's a bit long, but it goes through all the math). If you want to see the impact of size on sharpness check out this video: ruclips.net/video/VPzAAzEb1wE/видео.html
In short, there is an optimal value that you can calculate, but error on the side of smaller since the impact of diffraction isn't as bad as having a pinhole that is too large.
my website is at CyrusArthur.com
Let me know if you have any other questions
@@TheScienceofPhotography Thank you! I jumped around this playlist a bit and must have missed it. The math that you explain is really helpful.
Your lessons are just great. It really helped a lot. I have covered till here and will continue to watch. thanks a ton. :-)
Thank you!
Would curving the film plane eliminate or reduce vignetting, since the distance from the pinhole to the focal 'plane' would be similar at all points, (assuming an equidistant curve)? Thanks for the video.
Great question. Yes, if you match the curvature to the angle of projection from the pinhole then vignetting is greatly reduced. To eliminate it completely you would need a spherical film plane. However, the image will no longer be rectilinear and will have some level of distortion. Film cameras such as the widelux and the noblex have curved film planes and produce an almost fisheye effect. To reduce vignetting on a flat surface use a longer focal length (or pinhole distance) so that the difference in distance to the center and edges of the image is less. Check out my more recent pinhole videos on my 8x10 superwide camera to see the impact of vignetting. It is quite pronounced in my sample images.
Somewhat unrelated, but do you have any math on how to calculate how many pinholes you need for multi-pinhole imaging i.e. a collimator?
I’m not sure what I’m doing wrong. I have a pinhole diameter of .3 mm into thickness of .1 mm. When I use your formula I get a theta of 1.25 ?
I can’t imagine my viewing angle is 2.5°?
Looks like your calculator is giving an answer in radians since tan^-1(.3/.1) = 1.249 radians so the total angle is ~2.5 radians. Converting 2.5 radians into degrees is about 143 degree angle of view. Hope that helps!
I'm doing these math operations to know what's the angle of view that will be proyect on a wall of my bath room(yes, i want to use the window of my bath room as the plane of the pinhole...jajaja) so i have these parameters
f=1100mm
d=1.2mm
t=0.014mm(aluminium foil)
And all is equal to
Q=7,170.632701
What kind of number is that i mean what angle is that?
Thanks i really apreciate your lesson
You need to take the inverse tangent of d/t so when do you the calculation it equals 1.5591 radians or 89 degrees. Multiply that by 2 to get 178 degrees which is basically an unlimited angle of coverage. You will have an f/stop of f/916 which will make the image very dim so you would have to let your eyes adjust for a long time.
Why didn't you half d or t before you did your calculation? Shouldn't you have worked this instead?
tan-¹(.3/2)/(0.127/2)
I'm not sure I understand your question. I did have both d and t since I bought a pinhole of known diameter and known thickness. The expression you wrote in your comment simplifies to the one I used tan^-1(0.3/0.127) since both the numerator and denominator are divided by 2 it cancels out.
@@TheScienceofPhotography Sorry, it should have read half not have. I see what you mean about canceling out each other. My mistake. Like the vid.
@@henricvs No worries. Happy to clear things up.