On Thu, Feb 24, 2011 at 11:39 AM, John Sessoms <[email protected]> wrote:
> 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? There's an excellent astrophotographer named Thierry Legault, who photographs transits of the ISS in front of the sun. His webpage is full of information and photographs that will likely be of interest to you: http://www.astrophoto.fr/ He uses the CalSky.com website for planning. I went there, and an upcoming ISS pass at my location has an angular velocity of 0.52 deg/s. So we'll call it half a degree per second. Half a degree is the diameter of the full moon or the sun, BTW. Diagonal FOV for 500mm on APS-C is 3.3 deg. So the crossing time is ~6.3 seconds. > 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? Sure. > Where in the resulting image would the ISS be located? Pretty close to the corner where it started. Can't see it taking more than half a second to react and trip the shutter. > How many pixels would the ISS move across the FOV while the shutter was open > at the proposed 1/2500 exposure? This gets a little messy with the ISS crossing the frame diagonally (on a 3:2 sensor), since it's not moving parallel to a pixel edge, nor is it even moving along the diagonal of a square pixel. Let's work in units of "pixel edges" and not worry too much about a real pixel. The K-5 sensor is 4928x3264, so the diagonal is 5910 "pixel edges" in length, by the Pythagorean theorem. Crossing time was calculated to be 6.3 seconds, so the ISS crosses 5910/6.3 = 938 pixel-edges a second. Call it 1000 pixel-edges a second. So in 1/2500 s, it moves by less than half a pixel-edge. Not bad. > 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)? I don't see any reason it would, based on the calculations so far. -- PDML Pentax-Discuss Mail List [email protected] http://pdml.net/mailman/listinfo/pdml_pdml.net to UNSUBSCRIBE from the PDML, please visit the link directly above and follow the directions.
