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I was trying to calculate the diameter of Jupiter from a picture I took of it. Here's the information I was able to get that I needed to calculate the diameter:

Focal Length of Telescope: 1.2 m

Multiplier for Barlow Lens: 2x

Camera CCD (Sensor) dimensions: 14.8 x 22.2 (mm)

I found the diameter to be 135 pixels in my image. Here's what I did:

\begin{align} f_e &= \text{Focal Length $\cdot$ Barlow Multiplier}\\ &= 1.2 \cdot 2\\ &= 2.4 \end{align}

Then, to calculate the field of view:

$$\text{Field of View} = \frac{\text{Sensor Size}}{f_e} \cdot \frac{180}{\pi}$$

Where the sensor size is the length in millimetres of the size of the camera CCD (I used the horizontal axis, or the longer side, to get the 135 pixels, meaning that I'd use 22.2 mm), and the second fraction is there solely to convert radians into degrees. Substituting my values, I got:

\begin{align} \text{Field of View} &= \frac{0.0222}{2.4} \cdot \frac{180}{\pi}\\ &= 0.53^\circ \end{align}

As my entire image is 4271.8 pixels wide, and I know the angular field of view, I used ratios to get the field of view of the 135-pixelled Jupiter to be $\theta = 0.017^\circ$.

Then, using basic trigonometry, I was able to calculate the radius of jupiter:

$$\tan \left(\frac{\theta}{2}\right) = \frac{\text{Radius of Jupiter}}{\text{Distance to Jupiter}}$$

The photo was taken on March 22, 2016 - Wolfram Alpha gives 667.6 million as the distance to Jupiter on that day:

$$\tan \left(\frac{0.017^\circ}{2}\right) = \frac{\text{Radius of Jupiter}}{667.6 \cdot 10^6}$$

$$\therefore \text{Radius of Jupiter} = 97\ 577\ \mathrm{km}$$

Googling the radius of Jupiter gives $69\ 911\ \mathrm{km}$ as the result. Is there any reason why my value is so significantly off? Anything wrong with my procedure, anything I'm missing? I was not able to to upload this image I took as the file size was too large, however, the image does not lack clarity and I am completely stumped as to why my result is off by such a significant amount.

Any help will be greatly appreciated, thanks in advance.

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    $\begingroup$ Are the camera's pixels square? $\endgroup$ – Dr Chuck May 24 '16 at 6:57
  • $\begingroup$ Is all of the sensor illuminated? A stop in the optical path (like the focuser tube) can limit the effective fov. $\endgroup$ – Conrad Turner May 24 '16 at 12:45
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Your barlow is marked "2x" but I'll bet it isn't.

How is your camera mounted on the barlow? More importantly, what distance is between the camera's focal plane and the barlow?

An experiment to try (or at least to think about). Get an eyepiece extension tube (or improvise something) and put it between the camera and barlow, increasing the spacing. You will have to refocus a bit and you will get a different (I guess bigger) image size, even though the barlow is still "2x". To get an exact "2x" barlow you might need a particular spacing, and eyepieces are normally closer to the eyepiece than cameras because of the extra bulk of the camera. (I think this is the most likely cause of your discrepancy; your camera is currently mounted further back than an eyepiece would be so projection is making the image larger.)

A second experiment to try: without the barlow, photograph the moon and see how close you get to its expected size. If it's close, then our assumptions about the barlow are in question.

Third thing to try: are you absolutely sure about the distance from Earth to Jupiter? (It's probably OK but I'm always suspicious about online calculators.)

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A simpler approach would be to use a planetarium program to calculate the angular size of Jupiter at the date/time you want then use a bit of trig. to work out the angular size of each pixel of the camera

Tan(theta) = pixel size (in mm) over effective focal length in mm (2400mm)

(Sorry I don't know how to use the equation editor)

Then divide one by the other and you have the diameter of Jupiter in pixels.

Note: This assumes that the barlow is operating at 2x, as the barlow magnification factor is dependent on the distance of the camera sensor from the barlow's lens (and the focal length of the lens) this will probably not be accurate.

So if Jupiter is 38" in diameter and your pixel size is 5.6 microns 1 pixel = about 0.48" so Jupiter will be about 79 pixels in diameter.

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