From: Matthew Hunt
On Wed, Feb 23, 2011 at 10:24 PM, Darren Addy <[email protected]> wrote:
> When you are talking about an object that is only 47 arc seconds wide
> (at it's closest directly overhead) we would appear to be talking
> about an 8 pixel wide image.
(47 arcsec)*(0.500 m)/((5.48e-6m)*(206265 arcsec/radians)) = 20.7 pixels
(Where 5.48 microns = K-x pixel pitch)
Ok, since y'all are going to go all mathematical on me, some
hypotheticals ...
Suppose you just fix the camera on a tripod so it is pointed in a
direction where the ISS's orbit would cause it to pass through the Field
OF View[1] of the lens (as a diameter, not a chord) ... assume the
camera is mounted so that an image of the orbital path as focused on the
sensor by the lens approximates a diagonal of the sensor.
For the sake of argument, in this "experiment" the ISS will traverse the
frame as seen through the viewfinder from lower left to upper right.
Attach a 90 deg viewfinder if you have to, watch and trip the shutter as
soon as you recognize the ISS has entered the frame.
How long would it take for the ISS to traverse the FOV of the 500mm
mirror lens as seen through the viewfinder?
How long would it take to trip the shutter? i.e. would your reaction
time be quick enough to actually catch it within the frame?
Where in the resulting image would the ISS be located?
How many pixels would the ISS move across the FOV while the shutter was
open at the proposed 1/2500 exposure?
Does the combination of the rate at which the ISS is crossing the FOV
and rate at which the camera scans the sensor have any noticeable effect
(jelly shutter)?
[1] Maybe Field Of View is not the word I want ... but you know what I mean.
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