# Homework Help: Working out pixels

1. Nov 1, 2009

### greener1993

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I have the answer to all theses questions beacuse they are from a book but don't quite understand how to get it. Nice and good to cheat on my homework but not good in exam :S i am just so confused. i understand the resoultion is the smallest thing you can see in the picture which in most cause is a pixel which also in most cases is 1MM long. However i cant seem to gather how to work out questions. Why not use the book? beacuse i swear to god it is useless. OCR advancing physics book. Its more of a history book about physics than anything that will get us a good grade.

LS= light sensitive.
Red = Word I do not understand or there relationship to the question

1)a digital camera has a lens of focus length 50mm and produces an image of 1280x960 pixels. A human face (270x200)that is 1.5mfrom the lens fills the picture Of the light sensitive chip.
A) Estimate the dimension of the picture area of the LS chip in the camera.
B) Estimate the dimensions of one picture element in the LS chip.
C) estimate the scale of features on the face that can be resolved in the picture. could a person eyelashes be resolved?

2) Estimate how many pixels are needed in a satellite image of a town, if parked cars are to be resolved in the image.
Information : town = 10km2
lenth of car = 3m
resoulution = 3m

3)Estimate how many pixels are needed in an ultrasound scan image of a full-term foetus, if the baby’s fingers are to be distinguished
Information : Width of finger is 3mm
length of foetus a few hundred mm

4) Estimate the dimensions of an area of space imaged around a star, if an Earth-sized planet is just detectable in a 100 M pixel image.
Informtaion : Earth is about 10 000 km in diameter.

I just need it explaining well. Would be nice if you could include any equations and step by step instructions

Thank you :)

2. Nov 1, 2009

### jambaugh

2.) Suppose the entire town is one big parking lot. You would need to resolve each car right? Thus you need better than one pixel per car. How many cars fit in the town?

1.) involves http://en.wikipedia.org/wiki/Magnification" [Broken] for a given focal length and distance.

All the problems involve ratios and scaling. Draw some diagrams to help you conceptualize what's going on.

Last edited by a moderator: May 4, 2017
3. Nov 1, 2009

### greener1993

ive just work on 2 and 3 quickly would 2 be 33333333.33 or 3x10^8Pixels? and 3 100 pixels. Ive tryed to follow your info and draw pictures. If a 3m car can been seen then it would be 100,000 km.

4. Nov 1, 2009

### jambaugh

That last I can't quite follow cars and 100,000 km? 2.) Remember the pixel is an area so what area would it be to resolve a 3m car? Remember units of m^2. What again is the area of the town? 10km x 10 km = ? m^2.

Finally for 4.) remember the 100M pixel image is 100,000,000 pixel units of area (really 100x 1024x1024 since in computer speak 1K = 1023 not 1000)
What is the cross sectional area of the Earth relative to the area of space being observed?

It appears your instructor (or text) is trying to get you to think about the dimensions of what you are comparing. Remember scaling lengths by s scales areas by s^2 and volumes by s^3. That goes for unit conversions.

"Pixels" are tricky because they are variable units of area. So a 5foot resolution means a pixel area of 5ft by 5ft = 25 square feet per pixel.

If you systematically keep the units in the arithmetic and algebra you perform you will find it does much of the thinking for you and gives you confidence in your answers. Remember counts are unitless so to find number of pixels in an area (say 1000 square feet) find the area of one pixel (say 25 square feet per pixel) and when you divide all units need to cancel appropriately:

(1000 ft^2) / (25 ft^2/pixel) = (1000/25) pixels or 40 pixels.

Check it with 40 pixels times 25ft^2 per pixel yields 1000 ft^2.

That's about all the help I feel I should give you. Remember UNITS UNITS UNITS!!!

5. Nov 2, 2009

### greener1993

So it would be 3^2? so 9M per pixel? 10x10 =100m^2x 1000 = 100000m

100000 / 9?:S = 111111.1111 ? beacuse it say estimate. would it be 1x10^5? 9m seems huge for a pixel or is that the area it can see? maybe 100^2 / 3m^2? x1000? ..... im so confused i really am, im trying my best just cant get my head around it.

possibly, (100m^2)/(9m^2/pixel)= 11 pixels..... but the answer is 100 M pixels.

6. Nov 2, 2009

### greener1993

i cant do this at all its so confusing

7. Nov 2, 2009

### greener1993

Maybe (100km^2)/(1.732^2)?

I iwhs someone would tell me how to do them, theses arnt even homework questions no more, i have the answer im doing additional research beacuse i don't have a clue what im doing.

Im about ready to give up on this and say sod it ill get an F

If i dnt here from u again jambaugh, Thank you so much mate for trying to help me

8. Nov 2, 2009

### jambaugh

What is the area of the town in square meters (not square kilometers)

9. Nov 3, 2009

### greener1993

ive been doing it in meters......

10. Nov 3, 2009

### jambaugh

If the town is 10 km ^2 = 10 x 1km x 1km = 10 x 1000m x 1000m = 10,000,000 m^2.

Units Units Units.

You got the area per car-pixel is 9 m^2/pix

So (10,000,000 m^2)/(9 m^2/pix) = 1,111,111.111... pixels or 1.111...M pixel

You seem to be mixing the 100Mpixels of 4.) with this problem.

4.) If the Earth is about 10,000km in diameter then a square Earth cross section pixel is about 10,000 km x 10,000km = 100,000,000 km^2 per pixel.

Again work out the total area remembering to always incorporate your units in the math!
Units Units Units!!!

[Edit: More free advice. Go back and look at what mistakes you make. That is as important as seeing how to do it right. Your goal is to not do it wrong next time so it's about you and the process you use. Oh yea, and units, units units...]