Comment on what you estimate is your measurement accuracy. How many significant figures are meaningful? (In other words, can you measure the diameter to an accuracy of one pixel? One tenth of a pixel? One hundredth of a pixel? None of these?)
How much fainter in DN units is the sunspot than its surroundings?
Repeat your measurement of the Sun’s diameter several times and post your average value of the diameter in kilometers on the Blog. Include a comment about sig figs.
Describe the general nature of the sun's surface around the location of this sunspot.
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ReplyDeletethe most brightest DN is 4855 the fist lump of data on the histogram was the corona and the second one was of the inner sun.
ReplyDeleteit would be to one tenth of a pixel
960 pixels of length of the inner sun
958 pixels of length of the inner sun
967 pixels of the length of the inner sun
average is 961 DN
1089 pixels of length of the corona
units for the sun spot is 3500 down to 0 back to 3900 in 80 DN
the suns Dn is 3500 down to 0 also but after 989 DN
it is 961 pixels/Km
Step 2: The largest value I could find in the image of the sun was 5463.
ReplyDeleteStep 3: Average is 943 DN. Having 3 decimal places is unnecessary. The difference between measurements is much larger than hundredths or thousandths of a pixel, so these significant digits are meaningless. I think only tenths are necessary.
Step 4: The sunspot is about 1000 to 1500 DN units fainter than its surroundings.
Step 5: The average is ~ 13900000 km. The amount of significant figures are even less necessary now because the difference between measurements is a few hundred thousand.
Step 6: The sunspot itself is very dark, but the area immediately around it seems to be brighter than normal.
Assignment 2:
ReplyDeleteThe highest light value i could see inside the sun was 4899 ( the DN value is 4899)
The first lump of data on the histogram showed the corona of the sun. The second lump showed the inside of the sun. The full thing.
957, 957, 958, and 954 were the measurements ( length) of the sun that I got. I don't think that you need any decimal places because the length of the sun is a very large number and it is unnecessary. If you were measuring a smaller object and were using inches I would understand using decimals.
Yes you can. You can measure to the 10s of pixels but not hundredths of pixels.
The DN value of the sunspot is a third darker than the rest of the sun.
13628098, 13826725, 13873125 is the length in km.
Excellent work. If you have a diameter of 957 pixels for the diameter, this limits your final answer to 3 sig figs, so 13628098 would become 13600000 Km.
ReplyDeleteMy three measurements for the sun’s diameters are
ReplyDelete13910216
13965471.265
14023538.997
The average of the diameter is 13966408.74 km.
I found that changing the contrast of the sun actually changed the measurements by couple of pixels. If you use the Rainbow RGB, you can make pretty good accurate measurements. But the easiest one for me was the Spectrum contrast. I can make easy measurements using that because you can see the layers of the sun pretty clear. This was using the ImageJ.
And if you round that number, it would be 1390000000 km.
ReplyDeleteMy measurements were:
ReplyDelete13890164.860
13896217.235
13788238.489
I can round these numbers to three sigfig. Since for the pixels there are three significant figures, I would keep three significant figures on my measurements and leave the rest of the numbers zero like this: 13900000 Km