SITE AND ALL CONTENT COPYRIGHT@ 1999-2010 REMUS CHUA UNLESS OTHERWISE NOTED.
This site contains a personal collection of astronomy images taken with various astronomical instrument in selected locations around the world.
"Astronomers say the universe is finite, which is a comforting thought for those people who can't remember where they leave things."—Woody Allen
"With every passing hour our solar system comes forty-three thousand miles closer to globular cluster M13 in the constellation Hercules, and still there are some misfits who continue to insist that there is no such thing as progress."—Ransom K. Ferm.
SMEARING CALCULATIONS FOR PLANETARY ROTATIONS (FOR IMAGING PURPOSES)
Very often, AVI video sequence files are used to capture planetary images in the initial stage before using stacking software like Registax to stack the individual sub-frames of the AVI sequence. However, in order to maximize the amount of detail and features seen on the planetary surface, there is a limit to how long the AVI sequence should be, before planet rotationary smearing begins.
Once smearing occurs, details will be "blurred" and the end result will be an image that is not as sharp as one where the optimum smearing limit duration is not overlooked. Depending on the planet's rotation, we have varying limits for different planets, and each planet itself might have differentail rotational periods (for example Jupiter). In this case, we will assume the equatorial rotational speed of that planet instead. The following calculations will all be based on assuming the use of Phillips' Toucam Pro II camera which is widely used in the amateur astronomy imaging community.
CALCULATION EXAMPLE 1: MARS
Let us take Mars as our first example:
Assuming the Toucam Pro II web-camera is used in imaging (640x480 capture resolution, 3.87mm x 2.82mm chip dimension) and we are using an 8-inch SCT (2032 f.l.) with 5X Televue Powermate barlow:
Resolution (arc-seonds) / pixel = pixel size in microns of capture device / effective focal length (mm) of imaging instrument * 206
= 6/(2032*5)*206
= 0.12165" / pixel
Mars has an angular diameter as 20 arc seconds (assumption). It will therefore cover 20/0.12165 pixels on the Toucam imaging chip. This works out to 164.4 pixels. We take 164 pixels (take the lower rounded off limit) instead.
Next, we need to consider the linear rotation of Mars (km per minute) of its surface. Mars has a linear diameter of 6794km. Therefore it's circumference is
6794 Km diameter x 3.14159 = 21344km
The length of a typical day on Mars is 24.6597 hours. Therefore, the rotation per minute at the central meridian of Mars is
21344km / (24.6597 x 60 minutes) = 14.42km/min
As such, we are able to know now that each pixel spans a definite amount in km on Mars. For 164 pixels, each pixel at the central meridian spans:
6794km diameter / 164 = 41.43km
Finally, we are able to calculate the smearing rate (in minutes) per pixel.
(41.43km/pixel) / (14.42km/min) = 2.873 minutes/pixel
Therefore, it takes approximately 3 minutes for rotation to smear across one pixel at the central meridian where the effect is the worst. You can certainly use your own criteria as to how much smear is acceptable.
CALCULATION EXAMPLE 2: JUPITER
Let us now take Jupiter as our next example:
Again, we assume the same setup for the camera and imaging instrument as above, plus the same barlow:
Resolution (arc-seonds) / pixel = 0.12165" / pixel
However, Jupiter has a larger angular diameter of 40 arc seconds (assumption). It will therefore cover 40/0.12165 pixels on the Toucam imaging chip. This works out to 328.8 pixels. We take 328 pixels (take the lower rounded off limit) instead.
Next, we need to consider the equatorial linear rotation of Jupiter (km per minute) of its surface. The Jovian giant has a equatorial linear diameter of 142984km. Therefore it's circumference is:
142984km diameter x 3.14159 = 449197km
The length of a typical day on Jupiter is 9.9250 hours (based on System III 1965.0 sideral rotational period coordinates). Therefore, the rotation per minute at the central meridian of Jupiter is:
449197km / (9.9250 x 60 minutes) = 754.3km/min
As such, we are able to know now that each pixel spans a definite amount in km on Jupiter. For 328 pixels, each pixel at the central meridian spans:
142984km diameter / 328 = 435.9km
Finally, we are able to calculate the smearing rate (in minutes) per pixel:
(435.9km/pixel) / (754.3km/min) = 0.578 minutes/pixel
Therefore, it takes approximately 0.6 minutes (or 35 seconds) for rotation to smear across one pixel at the central meridian where the effect is the worst.
CALCULATION EXAMPLE 3: SATURN
Let us take Saturn as our last example:
Let us use the same imaging setup configuration again like above.
Resolution (arc-seonds) / pixel = 0.12165" / pixel
Saturn has an average angular diameter of 18.9 arc seconds (assumption). This is equivalent to 155 pixels. The ringed planet has a equatorial linear diameter of 119300km. Therefore it's circumference is:
119300km diameter x 3.14159 = 374792km
Saturn's equatorial rotational period is 10.2331 hours. Hence, the rotation per minute at the central meridian of Saturn is:
374792km / (10.2331 x 60 minutes) = 610.4km/min
As such, we are able to know now that each pixel spans a definite amount in km on Saturn. For 155 pixels, each pixel at the central meridian spans:
119300km diameter / 155 = 769.7km
Finally, we are able to calculate the smearing rate (in minutes) per pixel.
(769.7km/pixel) / (610.4km/min) = 1.26 minutes/pixel
Therefore, it takes approximately 1.3 minutes (or roughly 80 seconds) for rotation to smear across one pixel at the central meridian where the effect is the worst.