Luminous Landscape Forum
Equipment & Techniques => Digital Cameras & Shooting Techniques => Topic started by: AFairley on July 21, 2016, 12:42:47 pm
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Can any one tell me what the angle of view is of a horizontal panorama made by shifting a 17mm lens (such as the Canon) on a 24x36 sensor camera? Thanks.
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Can any one tell me what the angle of view is of a horizontal panorama made by shifting a 17mm lens (such as the Canon) on a 24x36 sensor camera? Thanks.
Hi Al,
On a 36mm wide sensor, with 2x 12mm shift added, it will create a virtual sensor width of 60mm at 17mm focal length.
That means that the approx. AOV is: 2x aTan((60/2) / 17) = 120.9 degrees at infinity focus.
Cheers,
Bart
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That means that the approx. AOV is: 2x aTan((60/2) / 17) = 120.9 degrees at infinity focus.
For horizontal AOV, yes, that's 120.9 degrees.
As to "equivalent FOV" (i.e. "what would be the FOV of the lens if measured as manufactures usually do, across the diagonal of the frame"), that would be
deg(2*atan((sqrt((60/2)**2 + (24/2)**2)) / 17)) = 124.5 degrees.
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For horizontal AOV, yes, that's 120.9 degrees.
As to "equivalent FOV" (i.e. "what would be the FOV of the lens if measured as manufactures usually do, across the diagonal of the frame"), that would be
deg(2*atan((sqrt((60/2)**2 + (24/2)**2)) / 17)) = 124.5 degrees.
And what is then an equivalent focal length for 124.5 degrees? And the same calculation for the 24 t/s, please?
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And what is then an equivalent focal length for 124.5 degrees?
One way of looking at it is to see what lens produces 124.5 degrees of FOV on 24x36mm sensor, which would give us
sqrt(24**2+36**2)/2 * tan(rad(90-(124.5/2))) = 11.38 mm
but that is, of couse, not entirely correct... The thing is that you'd have to crop the image from top and bottom to produce the same aspect ratio that "virtual 60x24" frame gives us... So we have to further shrink the short side of the frame to account for that "virtual extension" that shift gives us. So we get
sqrt((24*36/60)**2+36**2)/2 * tan(rad(90-(124.5/2))) = 10.2 mm
All in all, it means that you can produce similarly-looking frame by snapping 10mm hyper-wide Voigtlander and then cropping the image to 60:24 == 5:2 aspect ratio.
And the same calculation for the 24 t/s, please?
By how many millimeters it shifts?
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By how many millimeters it shifts?
Looked it up... According to DPR (https://www.dpreview.com/reviews/canon-24-3p5-tse-c10) it's those same 12mm.
So, equivalent FOV:
deg(2*atan((sqrt((60/2)**2 + (24/2)**2)) / 24)) = 106.79 degrees
Equivalent focal length on FF sensor:
sqrt(24**2+36**2)/2 * tan(rad(90-(106.79/2))) = 16.07 mm
Equivalent focal length on FF sensor if cropped to the same aspect ratio:
sqrt((24*36/60)**2+36**2)/2 * tan(rad(90-(106.79/2))) = 14.4 mm
That is, Canon's own 14mm prime would do.
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Can any one tell me what the angle of view is of a horizontal panorama made by shifting a 17mm lens (such as the Canon) on a 24x36 sensor camera? Thanks.
Towards the end of this page http://www.cambridgeincolour.com/tutorials/tilt-shift-lenses1.htm there is a convenient shift calculator. You can even select camera orientation and shift direction.
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Towards the end of this page http://www.cambridgeincolour.com/tutorials/tilt-shift-lenses1.htm there is a convenient shift calculator. You can even select camera orientation and shift direction.
Interesting link, thanks.
Doesn't mention diagonal FOVs (am I the only one who tend to think about lenses and images they produce in terms of diagonal FOV?), but at least the numbers those guys arrived at for focal lengths necessary to produce equivalent (including aspect ratio) images on FF without shifts match mine (10.2 and 14.4), which is always good to know...
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Thanks, Bart and George, horizontal FOV was the number for me as that describes where the edges of the image will be.
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You don't really need to use a complicated calculation for figuring out the new FL equivalent when shifting, just use the ratio of the FF vs new frame with shifts, egs;
12mm left and right shift on FF using a 17mm lens
FF width = 36mm
Shifted width is 36 + 12 + 12 = 60mm
New FL equiv is 17 * 36/60 = 10.2mm
12mm left and right shift on FF using a 24mm lens
24 * 36/60 = 14.4mm
10mm left and right shift on FF using a 28mm lens
New frame width is 36 + 10 + 10 = 56mm
28 * 36/56 = 18mm
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You don't really need to use a complicated calculation for figuring out the new FL equivalent when shifting, just use the ratio of the FF vs new frame with shifts, egs;
12mm left and right shift on FF using a 17mm lens
FF width = 36mm
Shifted width is 36 + 12 + 12 = 60mm
New FL equiv is 17 * 36/60 = 10.2mm
12mm left and right shift on FF using a 24mm lens
24 * 36/60 = 14.4mm
10mm left and right shift on FF using a 28mm lens
New frame width is 36 + 10 + 10 = 56mm
28 * 36/56 = 18mm
Indeed.
But it's much less fun that way, isn't it? ;)
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You don't really need to use a complicated calculation...
Damn, Mike, you are entering a dangerous territory. You remind me of a medieval passerby, overhearing scholars discussing how many teeth a horse might have and suggesting to simply open the horse's mouth and count. They beat the crap out of him for being un-scientific :D
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You remind me of a medieval passerby, overhearing scholars discussing how many teeth a horse might have
Yeah, those were the days... <sigh>
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:D ;D