How big is the comet based on a 735x731px surface picture taken from 3km away?

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The discussion centers on calculating the width of a comet based on a 735x731 pixel image taken from 3 kilometers away, with the camera's resolution being 3.5 times lower than Tycho Brahe's. The initial calculation suggested a width of 175 meters, which participants deemed too small, estimating the comet's actual width to be around 2.2 kilometers. Key points include the need to convert the angular size of each pixel (3.5 arc minutes) into radians to determine the linear size at the given distance. Participants emphasize using geometric principles, such as the tangent function, to refine the calculations. The conversation highlights the importance of understanding angular measurements in relation to distance for accurate size estimation.
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Homework Statement


There is a picture, size 735x731px, that show the surface of comet that was taken from a probe distanced 3 kilometers from the comet. If the resolution of the camera that recorded the surface is 3.5 times lower then the resolution of Tycho Brache (that is 1 angular minute, 1/60 of a degree), what is the width of the comet? Use expresion for α (angle), arch (l) and distance (r)

Homework Equations


α=l/r
l=α*r

The Attempt at a Solution


l=(3.5/60)*3000meters
l=175meters
To me this seems to small, the comet is like 2,2km wide or more. Also I have no idea in what way to use the pixels to calculate the right size.
Any help would be nice
 
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Hi Boxblax, Welcome to Physics Forums.

Boxblax said:

Homework Statement


There is a picture, size 735x731px, that show the surface of comet that was taken from a probe distanced 3 kilometers from the comet. If the resolution of the camera that recorded the surface is 3.5 times lower then the resolution of Tycho Brache (that is 1 angular minute, 1/60 of a degree), what is the width of the comet? Use expresion for α (angle), arch (l) and distance (r)

Homework Equations


α=l/r
l=α*r

The Attempt at a Solution


l=(3.5/60)*3000meters
l=175meters
To me this seems to small, the comet is like 2,2km wide or more. Also I have no idea in what way to use the pixels to calculate the right size.
Any help would be nice
Wouldn't the resolution information imply that each image pixel represents angular size of 3.5 arc minutes? How many pixels across is the picture?
 
It is 731px across. Could you please explain more how to solve it
 
If each pixel represents an angular size of 3.5 arc minutes, the what linear size does each pixel represent at 3 km distance?
 
Yes, that is what i need to find out. I need to find out the real size of comet. How to find out the size of 1 pixel in meters?
 
Boxblax said:
Yes, that is what i need to find out. I need to find out the real size of comet. How to find out the size of 1 pixel in meters?
Draw it out. You have an angle of 3.5 arc minutes and a range of 3 km. What's the length of the arc on the end of that radius? Look at your Relevant equations.
 
175meters? I don't quite understand what you want to do. Could you solve it?
 
Helpers are not allowed to solve homework problems for you. We can only provide hints and suggestions, or point out errors. It's in the forum rules (which you read right? :smile:).

175 meters is not correct. One pixel spans an angle of 3.5 arc minutes. That's 3.5 x (1/60) degrees. With a radius of 3 km, what is the length of the arc that is swept out by that angle? Show your calculations.
 
my calculations are in a photo. still doesn't seem right to me, maybe i am wrong
 

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  • #10
Remember that geometrical calculations involving angles should use radians. Otherwise, you can approximate the figure with a right angle triangle since the radius is so large, and use the tangent function.

I'd convert the angle in arcminutes to radians in your calculation.
 
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thanks mate
 

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