# Calculating the Diameter of Jupiter through Image [closed]

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.

• It appears that your sensor is of the Canon APS-C type which you have assumed has about 12 mega pixels? This gave you by calculation? your 4271.8 pixel width. Did you actually count the number of pixels across the field of view? A quick way of finding the width of Jupiter relative to the width of the frame is to just measure the distances on an enlarged print. May 22, 2016 at 16:17
• Using the known values for radius and distance to, you can determine that $\theta \sim 0.012^{\circ}$. The ratio of your CCD dimensions (i.e., ~0.6667) nearly make up the difference, but I am not sure if that is a coincidence. May 22, 2016 at 17:46
• @Farcher I used gimp to get the coordinates of various points on the image, including that of the two points on the circumference of Jupiter (the x-coordinates of which I then subtracted to get the pixel diameter of Jupiter) as well as the width of the image in pixels. Where did I assume that my sensor has about 12 mega pixels? Didn't I find the width of Jupiter relative to the width of the frame already by dividing the 135 pixels by the 4271.8 (right after I calculated the field of view of the entire image)? May 22, 2016 at 22:58
• It was because your number of pixels was fractional that I assumed that you had done it by calculation rather than "counting". As what you have done seems logical I was looking for a possible reason for the large discrepancy as it appears was @honeste_vivere. May 22, 2016 at 23:18