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

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pchaplo

pchaplo

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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 ...'
 
Hmmm....let see. Your flying a plane and taking pictures at the same time. Do you have a co-pilot with you....cause that sounds kind of dangerous even with all the autopilots these days.

Also, what aircraft are you flying and how tall is it?
pchaplo said:
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 ...'
Click to expand...
--

-------------------------------------------------------------------------------------------------
My little test site for right now:
http://homepage.mac.com/roiegat/Misc/PhotoAlbum71.html
 
What you want to do is get a really big ruler and lay it on the ground then take a picture of it from various heights to work out what height you need to be at to get the 1.5 miles in the horizontal. You can also check for backfocus at the same time :).

(The ruler idea isn't as crackpot as it may seem, a known distance measured out on the ground could be used in a similar way to the scale on a map. e.g fly over a running track with a 100m or yard marking easily visible from the sky when you are flying over it, then see how many of them fit into your picture at a specific height, shouldn't take more than 2 or three shots to allow you to calibrate your height properly. Might be more accurate than relying on lens characteristics and ratios)
pchaplo said:
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 ...'
Click to expand...
 
"Fly" in the general sense of "flying the assignment" I use a professional pliot when Im working. Shooting keeps my hands full.

.... and the plane is red, if that helps :-) LOL!

-Paul
roiegat said:
Also, what aircraft are you flying and how tall is it?
pchaplo said:
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 ...'
Click to expand...
--

-------------------------------------------------------------------------------------------------
My little test site for right now:
http://homepage.mac.com/roiegat/Misc/PhotoAlbum71.html
Click to expand...
--
'... they need no candle, neither the light of the sun ...'
 
I actually wasn't kidding about the height of the plane. You see the device that measures altitude is typically located in the front portion of the plane. If this device is lets say....5 feet off the ground...you need to account for that in the calculation.

It's good to hear you got a pilot. I also like red planes. I think it would be cool to do aerial photography....
pchaplo said:
.... and the plane is red, if that helps :-) LOL!

-Paul
pchaplo said:
pchaplo said:
Also, what aircraft are you flying and how tall is it?
Click to expand...
Click to expand...
Click to expand...
\
 
George,

LOL on the backfocus - should the road be at an angle? Can the propeller be adjusted by Canon? ;-)

The giant ruler works with known landmarks, etc. Right - I do that locally, but sometimes I dont have the luxury & the scale changes. I know that it can be calculated. Id like to learn how.

Thank you,
Paul
George Eliott said:
(The ruler idea isn't as crackpot as it may seem, a known distance
measured out on the ground could be used in a similar way to the
scale on a map. e.g fly over a running track with a 100m or yard
marking easily visible from the sky when you are flying over it,
then see how many of them fit into your picture at a specific
height, shouldn't take more than 2 or three shots to allow you to
calibrate your height properly. Might be more accurate than relying
on lens characteristics and ratios)
pchaplo said:
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 ...'
Click to expand...
Click to expand...
--
'... they need no candle, neither the light of the sun ...'
 
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
 
Sounds like you've done this thing before.

Easiest answer.....as you state, its a crop factor (for field of view factor). You can probably do some reverse math. To fit 1.5 miles of land onto the sensor, you need to capture (1.5 * 1.6 =) 2.4 miles of land.

To calc, you have two triangles, exactly the same. Right angle. The "bottom" will be 1.2 miles, and the angle of the opposing side is 32*

Use the fomula to calc the height. (I've stretched my mind back to high school as much as I can)

(Not sure if the forum will completly mess up this diagram. Basically draw a straight vertical line, it is 2.4 miles long

Draw a horizontal line out. This is the distance you need to determine

Join the ends, to form two triangles. The total angle there is 64*, or 32* for each right angled triangle.

. ---
. ~ ---
. ~ ---
2.4m -------------- Total 64*, so each triangle = 32* at this point
. ~ ---
. ~ ---
. ---
pchaplo said:
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 ...'
Click to expand...
 
You only need to do it once for a particular lens camera combination, whatever calibration system you go for you need to know your height above the ground you are measuring, don't know if an altimeter will give you an accurate enough reading or not.
pchaplo said:
LOL on the backfocus - should the road be at an angle? Can the
propeller be adjusted by Canon? ;-)

The giant ruler works with known landmarks, etc. Right - I do that
locally, but sometimes I dont have the luxury & the scale changes.
I know that it can be calculated. Id like to learn how.

Thank you,
Paul
George Eliott said:
(The ruler idea isn't as crackpot as it may seem, a known distance
measured out on the ground could be used in a similar way to the
scale on a map. e.g fly over a running track with a 100m or yard
marking easily visible from the sky when you are flying over it,
then see how many of them fit into your picture at a specific
height, shouldn't take more than 2 or three shots to allow you to
calibrate your height properly. Might be more accurate than relying
on lens characteristics and ratios)
pchaplo said:
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 ...'
Click to expand...
Click to expand...
--
'... they need no candle, neither the light of the sun ...'
Click to expand...
 
I'm not sure that the crop factor has anything to do with, nor the focal length. The field of view is 64 deg. Using the right angle rule, let's assume the angle is 1/2 of 64, or 32 deg. The length along the ground is 0.75 miles or 3960 ft. Tangent = Opposite/Adjancent. Opposite is the 3960 ft and Adjacent is your height. Therefore Height = Opp/(Tan 32); = (3960/0.5773) = 6859 ft.

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.
pchaplo said:
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 ...'
Click to expand...
 
bobvj said:
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.
pchaplo said:
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 ...'
Click to expand...
Click to expand...
--
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.
 
I punched in 30 deg not 32 deg. The tan of 32 is 0.625.

Corrected answer for the height is therefore 3960/0.625 = 6337 Ft.

By flying 500 ft higher, at 6837, you will get 6837*0.625 = 4273 ft, and then doubling to get approx 8550 ft along the ground.
bobvj said:
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.
pchaplo said:
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 ...'
Click to expand...
Click to expand...
 
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? 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!

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

Great Light!
Paul
Doug Kerr said:
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
Click to expand...
--
'... they need no candle, neither the light of the sun ...'
 
pchaplo said:
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 ...'
Click to expand...
 
Not according to the formula that Doug gave you, looks like doug made a msitake in applying the formula by multplying W by w/F instead of F/w
pchaplo said:
7920 ft. = 22.7mm (D/28mm)
D=9769 ft.

Paul
Doug Kerr said:
pchaplo said:
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 ...'
Click to expand...
Click to expand...
--
'... they need no candle, neither the light of the sun ...'
Click to expand...
 
The trig approach yielded 6500 ft. while the "Width of object field" yielded 9700 - did I mess up with the W=w(D/F) calculations? :

7920 ft. = 22.7mm (D/28mm)
D=9769 ft.

Paul
Doug Kerr said:
pchaplo said:
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 ...'
Click to expand...
Click to expand...
--
'... they need no candle, neither the light of the sun ...'
 
Based on my earlier post, try this test.

Use the side of a house or a long wall.

Measure 20 ft along the wall with both ends marked.

Find the mid point (10 ft) and back your camea away perpendicular until both end points are on the edge of the viewfinder. You should be 16 ft away from the wall at that point.

Tan 32 = 0.625 (1/2 of 64 deg lens field)

Tan = Opp/ Adj where Opp = 10 ft and ADJ = distance from wall

ADJ = 10/0.625 = 16ft
 
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
bobvj said:
Corrected answer for the height is therefore 3960/0.625 = 6337 Ft.

By flying 500 ft higher, at 6837, you will get 6837*0.625 = 4273
ft, and then doubling to get approx 8550 ft along the ground.
bobvj said:
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.
pchaplo said:
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 ...'
Click to expand...
Click to expand...
Click to expand...
--
'... they need no candle, neither the light of the sun ...'
 
Doug Kerr said:
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
Click to expand...
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
 
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
pchaplo said:
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
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