Angle/width of view with 1.6 crop factor (math whiz's welcome)

[George Eliott]

A similar but easier to do method than my running track idea, one point to remember though is that the viewfinder doesn't give 100% coverage you will need to take a picture to confirm that the dots are at the edges of the frame.

Bob

Bob,

Thanks! I was wondering: it seems like we could accomodate the
"crop factor" of the sensor by either adjusting the angle of view,
or perhaps the length along the ground. The crop factor implies
that the angle of view is narrower, right?

Does crop factor mean that the horizontal and vertical dimensions
of the frame are decreased by a factor of 1.6? For example: is it
correct to adjust the "target" length on the ground like this:

1.5 miles = 7920 ft.
7920 ft. * 1.6 = 12672 ft. [wow!] as the adjusted target length?
Then apply tangent formula?

Or is the 1.6 crop factor for the area of the image and/or is
this all proportional?

Thanks,
Paul

 

[Doug Kerr]

Hi, Paul,

Doug,

Thanks again - you are THE man!

W=w(D/F) "(sensor is explicitly treated)"

Thanks for giving me the formula, but not spoon feeding:

7920 ft. = 22.7mm (D/28mm) [note: right: mm/mm, very nice!]

D=9769 ft. Ja?

Jawohl. (But somehow I got another number I posted in the subject line later - not sure what I did wrong. Your algebra is better than mine!)

This is very useful. Maybe I want to shoot wider...

Also, I can have a "factor" for my sensor/ fixed f.length lens
combination (.811) that is handy!

Of course.

PS: Keep up the reminders on words & correct usage - we all need
those reminders & it does make a difference.

Thanks.

Best regards,

Doug

 

[Doug Kerr]

Hi, Paul,

I had used a 1-mile wide object field when I got that other number! Duh! It's so hard to get good help anymore!

Best regards,

Doug

 

[Doug Kerr]

Sorry - that result was for an object field 1.0 miles wide.

Best regards,

Doug

 

[pchaplo]

Thanks, Nico.

Paul,

If the subject distance is large compared to the focal length
(certainly true in the case you describe) the width of the "object
field" (the width of the area in real space whose image fills the
width of the frame) can be calculated thus:

W = w (D/F)

where W is the width of the object field, w is the width of the
camera format (22.7 mm for the 300D), D is the object distance of
interest, and F is the focal length of the lens.

Regarding units: if w and F are in a consistent unit (mm, for
example), then we can treat D and W in a different consistent unit
(feet, for example). How handy!

Beacuse the size of the camera format (sensor is explicitly
treated), there is no need to take any format size factor into
account.

Best regards,

Doug

To be precise, the formula is W=w*((D/F)-1). That way the formula
works for close-up/macro work.

In our case D/F is a lot larger than 1, so (D/F)-1 can be
approximated by D/F.

Nico

--
'... they need no candle, neither the light of the sun ...'

 

[pchaplo]

Sorry - that result was for an object field 1.0 miles wide.

Best regards,

Doug

--
'... they need no candle, neither the light of the sun ...'

 

[bobvj]

After you get the distance from the test below, move back an additional 5 ft. Take your picture, find how far the end points are from each other now and multiply by 100,

Paul,
Unfortunately I'm an engineer, and not a camera expert. I really
don't know what effect the crop factor has. It just seems to me
that the lens determines the angle of view.
I've suggested a test below and maybe you can use a modified
version. You want 7920 feet. Let's factor everything by 1/100.
Find a wall and mark 2 points 79 ft apart. Back you camera away
perpendicularly from the midpoint(39.5 ft) until both end points
are in your viewfinder. ( I think this will be about 63 ft away).
Whatever the distance is, you'll need to fly 100 times higher than
that reading.
We can discuss formulas all day, but a simple test is the way to be
confident.

Bob

 

[MacNico]

I am trying to determine how high I need to fly to fit 1.5 miles of
land in the longest dimension (horizontal) of a 300D frame. The
view will be straight down, perpendicular to the ground.

Here are the givens:

Lens: 28mm Nikon PC lens on Canon 300D body. Angle of view max.
(original specs from Nikon): 74 degrees, or 64 degree (horizontal)
the latter seems more usable. (I believe that this is diagonal on a
35mm FF camera - does not apply to the smaller digital sensor size).

Im in the US, so I want to use feet: 1. mile = 5280 feet; 1.5
mi=7920 ft.
The trick is: how to adjust for the crop factor, and then use
geometry to get 7920 ft. in the horizontal dimension of the frame.
Or another approach?

Remember: for full credit, you must "show your work" : ) Just
kidding! Actually, I want to learn how to do this, so please show
me how?

Extra credit: if I fly 500 ft. higher than the above answer, how
much more of a view (in feet) do I get?

Disclaimer: this is not homework, Im really flying this!

Thanks,
Paul

--
'... they need no candle, neither the light of the sun ...'

So are we all in agreement that the altitude required is in 9500-10000ft ballpark no?

Nico

 

[dwalby]

If I'm not mistaken, when calculating lens effective viewing angle, you use the diagonal of the frame, not the longest side. In other words, you need a circle of light large enough to completely cover the entire frame, which would be a diameter equal to the diagonal.

I'm at work, and don't feel like looking it up, but if you do a quick calculation with the diagonal of the Rebel sensor and the 18 or 55 mm focal lengths of the kit lens, I think you get the claimed viewing angle of the lens according to what's specified in the user manual. Or, try it with a typical 35mm lens and use the diagonal of 43.2666mm rather than the long end 36mm.

The lens angle of view is 2*arctangent(diagonal/(2*focal length)).

I agree with your formula otherwise.

Paul,

If the subject distance is large compared to the focal length
(certainly true in the case you describe) the width of the "object
field" (the width of the area in real space whose image fills the
width of the frame) can be calculated thus:

W = w (D/F)

where W is the width of the object field, w is the width of the
camera format (22.7 mm for the 300D), D is the object distance of
interest, and F is the focal length of the lens.

Regarding units: if w and F are in a consistent unit (mm, for
example), then we can treat D and W in a different consistent unit
(feet, for example). How handy!

Beacuse the size of the camera format (sensor is explicitly
treated), there is no need to take any format size factor into
account.

Best regards,

Doug

 

[dwalby]

not in agreement, see my other post above regarding using the sensor diagonal length not the width.

Using the diagonal (27.2635mm) I get 8134 ft.

So are we all in agreement that the altitude required is in
9500-10000ft ballpark no?

Nico

 

[dwalby]

While its true the diagonal is what should be used to calculate the nominal angle of view for a particular focal length lens, I forgot to take into account the fact that the circle projected by the lens is greater than the horizontal dimension of the sensor (or film).

So, for calculating the distance, the 22.7mm horizontal dimension of the sensor is what gets scaled, not the diagonal of the frame. I am in agreement with the rest of the crowd. The height calculates to 9769 ft. using the horizontal dimension of the sensor.

sorry for the confusion

Using the diagonal (27.2635mm) I get 8134 ft.

So are we all in agreement that the altitude required is in
9500-10000ft ballpark no?

Nico

 

[dwalby]

While its true the diagonal is what should be used to calculate the nominal angle of view for a particular focal length lens, I forgot to take into account the fact that the circle projected by the lens is greater than the horizontal dimension of the sensor (or film).

So, for calculating the distance, the 22.7mm horizontal dimension of the sensor is what gets scaled, not the diagonal of the frame. I am in agreement with the rest of the crowd. The height calculates to 9769 ft. using the horizontal dimension of the sensor.

sorry for the confusion

I'm at work, and don't feel like looking it up, but if you do a
quick calculation with the diagonal of the Rebel sensor and the 18
or 55 mm focal lengths of the kit lens, I think you get the claimed
viewing angle of the lens according to what's specified in the user
manual. Or, try it with a typical 35mm lens and use the diagonal
of 43.2666mm rather than the long end 36mm.

The lens angle of view is 2*arctangent(diagonal/(2*focal length)).

I agree with your formula otherwise.

Paul,

If the subject distance is large compared to the focal length
(certainly true in the case you describe) the width of the "object
field" (the width of the area in real space whose image fills the
width of the frame) can be calculated thus:

W = w (D/F)

where W is the width of the object field, w is the width of the
camera format (22.7 mm for the 300D), D is the object distance of
interest, and F is the focal length of the lens.

Regarding units: if w and F are in a consistent unit (mm, for
example), then we can treat D and W in a different consistent unit
(feet, for example). How handy!

Beacuse the size of the camera format (sensor is explicitly
treated), there is no need to take any format size factor into
account.

Best regards,

Doug

 

[pchaplo]

If you add 500 ft to height, you are at 7359 ft. Opp =
7359*0.5773=4248 ft which need to be doubled. That's approx 9500
ft.

I am trying to determine how high I need to fly to fit 1.5 miles of
land in the longest dimension (horizontal) of a 300D frame. The
view will be straight down, perpendicular to the ground.

Here are the givens:

Lens: 28mm Nikon PC lens on Canon 300D body. Angle of view max.
(original specs from Nikon): 74 degrees, or 64 degree (horizontal)
the latter seems more usable. (I believe that this is diagonal on a
35mm FF camera - does not apply to the smaller digital sensor size).

Im in the US, so I want to use feet: 1. mile = 5280 feet; 1.5
mi=7920 ft.
The trick is: how to adjust for the crop factor, and then use
geometry to get 7920 ft. in the horizontal dimension of the frame.
Or another approach?

Remember: for full credit, you must "show your work" : ) Just
kidding! Actually, I want to learn how to do this, so please show
me how?

Extra credit: if I fly 500 ft. higher than the above answer, how
much more of a view (in feet) do I get?

Disclaimer: this is not homework, Im really flying this!

Thanks,
Paul

--
'... they need no candle, neither the light of the sun ...'

--
fredyr
May Day, May Day! calls pilot to control tower. Reply. Give us your
height and your position. Answer. I am 5 foot 2 and sitting in the
front!
Remembered from the University of New South Wales recorded list of
jokes. Have a nice day.

--
'... they need no candle, neither the light of the sun ...'

 

[pchaplo]

Paul,
Unfortunately I'm an engineer, and not a camera expert. I really
don't know what effect the crop factor has. It just seems to me
that the lens determines the angle of view.
I've suggested a test below and maybe you can use a modified
version. You want 7920 feet. Let's factor everything by 1/100.
Find a wall and mark 2 points 79 ft apart. Back you camera away
perpendicularly from the midpoint(39.5 ft) until both end points
are in your viewfinder. ( I think this will be about 63 ft away).
Whatever the distance is, you'll need to fly 100 times higher than
that reading.
We can discuss formulas all day, but a simple test is the way to be
confident.

Bob

--
'... they need no candle, neither the light of the sun ...'

 

[pchaplo]

So, for calculating the distance, the 22.7mm horizontal dimension
of the sensor is what gets scaled, not the diagonal of the frame.
I am in agreement with the rest of the crowd. The height
calculates to 9769 ft. using the horizontal dimension of the sensor.

sorry for the confusion

Using the diagonal (27.2635mm) I get 8134 ft.

So are we all in agreement that the altitude required is in
9500-10000ft ballpark no?

Nico

--
'... they need no candle, neither the light of the sun ...'