Calculating Video Recording Capability of 4500 ft Object w/ Camcorder

[Max CR]

I am working on this as an outside of school project.

An object is 4500 feet away from the video camera. Will my video camera be able to video record it? I have no idea how to calculate this, and I seriously need some help in understanding this. Please include any math equations you use to figure this out since I need to do this again with other cameras.

Object specifications:
3.5 inches by 65.5 inches

Camera specifications:
42X optical zoom lens

Camcorder Sensor Resolution 800 Kpix

Camcorder Effective Video Resolution 380 Kpix

Camcorder Effective Still Resolution 0.3 Mpix

Optical Sensor Type CCD

Optical Sensor Size 1/6"

Min Illumination 2 lux

Digital Zoom 2000 x

Max Shutter Speed 1/8000 sec

Min Shutter Speed 1/30 sec

Image Storage JPEG 640 x 480 ¦ JPEG 640 x 360

Type Zoom lens - 1.8 mm - 75.6 mm - f/1.9-4.3

Focal Length 1.8 mm - 75.6 mm

Focus Adjustment Automatic, manual

Lens Aperture F/1.9-4.3

Optical Zoom 42 x

Zoom Adjustment Motorized drive

Features Built-in lens shield

Low Lux / Night Mode Yes

Digital Still Camera Function Yes

Type LCD display - TFT active matrix - 2.7" - color

Display Form Factor Rotating

Display Format 123,200 pixels

The link where I got this information is at:
http://accessories.us.dell.com/sna/productdetail.aspx?sku=A1538817&cs=19&c=us&l=en&dgc=SS&cid=27530&lid=627063

 

[Pengwuino]

What do you mean be able to record it? How small is the object? If it's a mountain, I don't think you'll have a problem...

 

[negitron]

Honestly, this is one of those things you can VERY easily determine experimentally. Calculations are all well and good but nothing beats actually seeing the result.

 

[rcgldr]

You can calculate the subtended angle of the 3.5 inch (3.5 / 12 foot) width:

http://en.wikipedia.org/wiki/Visual_angle

In this case, angle = tan^-1(3.5 / (2 x 12 x 4500)) = .001857 degrees, not much.

 

[russ_watters]

Honestly, this is one of those things you can VERY easily determine experimentally. Calculations are all well and good but nothing beats actually seeing the result.

Here's how: take a video of an object of known size (such as a ruler) at a known distance, then count how many pixels across it is. Use trigonometry to calculate the resolution in arcsec per pixel, then compare that resolution to the angular size of the object provided by Jeff.

 

[Max CR]

Here's how: take a video of an object of known size (such as a ruler) at a known distance, then count how many pixels across it is. Use trigonometry to calculate the resolution in arcsec per pixel, then compare that resolution to the angular size of the object provided by Jeff.

What do you mean by "Use trigonometry to calculate the resolution in arcsec per pixel, then compare that resolution to the angular size of the object provided by Jeff." This is where I got lost.

THanks to everyone else. I have read your posts.

 

[Max CR]

You can calculate the subtended angle of the 3.5 inch (3.5 / 12 foot) width:

http://en.wikipedia.org/wiki/Visual_angle

In this case, angle = tan^-1(3.5 / (2 x 12 x 4500)) = .001857 degrees, not much.

Where did you get these numbers and formaulas from? Please show your work. Thanks

 

[rcgldr]

You can calculate the subtended angle of the 3.5 inch (3.5 / 12 foot) width:

http://en.wikipedia.org/wiki/Visual_angle

In this case, angle = tan^-1(3.5 / (2 x 12 x 4500)) = .001857 degrees, not much.

Where did you get these numbers and formulas from?

From the wiki article. Imagine a right triangle where the hypotenuse goes from the lens to one edge of the 3.5 inch wide object. The adjacent side is the distance from the lens to the center of the object, and the opposite side is goes from the center of the object to the edge, or 1.75 inches (3.5/2) in this case. The adjacent side of the triangle is 4500 feet, and the opposite side is .14583333 feet (1.75/12). The tangent is (1.75/12)/4500.

 

[Phrak]

Unfortunately for that formula Jeff, we don't know that the veiwing field at a distance of 1 foot is 3.5 inches, or did I miss something in the camera specs?

Max CR, do you actually have this camera on hand, so you can experiment with it, as suggested? If so, you don't need curious formula, just comparison of similar triangles and a measuring tape or ruler.

 

[rcgldr]

Unfortunately for that formula Jeff, we don't know that the veiwing field at a distance of 1 foot is 3.5 inches, or did I miss something in the camera specs?

3.5 inches is the stated width of the target object.

 

[Redbelly98]

An object is 4500 feet away from the video camera. Will my video camera be able to video record it? I have no idea how to calculate this, and I seriously need some help in understanding this. Please include any math equations you use to figure this out since I need to do this again with other cameras.

To answer this question, calculate the following; if the result is larger than 1, your camera should be able to see the object:

( D_t / L_t ) × N_t × ( L / D )​

where

L_t and D_t are the length of and distance away, respectively, of a nearby test object. Note, D_t should be at least 3 times larger than L_t.
N_t is the number of pixels that the test object takes up in the image, along it's entire length.
D and L are the distance away and length of, respectively, of the object of interest.
​

In your case, D is 4500×12 inches, and L is 3.5 inches, so L/D is 1/15,400.

 

[Max CR]

Jeff, thanks for posting that link. I reviewed your work and I believe I understand it, but red belly's formula seems much easier to work with. I am going to try and see if I can use her formula and if there is no success I will try yours. Thank you. Phrak, I am going to experiment with it using Red's formula.

Red, the only thing that I would like to know is why aren't we using meters in this formula since it is standard notation? Why are we using inches like you suggested?

 

[negitron]

What units did you use in your OP?

 

[Redbelly98]

Negitron answered it. But to elaborate a little ...

As long as you use the same units for Dt and Lt, and also use the same units for D and L, the formula will work. The ratios in the formula, Dt/Lt and L/D, will have the same values no matter what units are used.

You could even use inches for D and L, and meters for Dt and Lt.

Note, L is the smaller dimension of the 3.5" x 65" object size.

 

[rcgldr]

You can calculate the subtended angle of the 3.5 inch (3.5 / 12 foot) width:
http://en.wikipedia.org/wiki/Visual_angle
In this case, angle = tan^-1(3.5 / (2 x 12 x 4500)) = .001857 degrees, not much.

Correction, angle = 2 x tan^-1(3.5 / (2 x 12 x 4500)) = 0.0037136 degrees

Jeff, thanks for posting that link.

Don't forget to include the optical zoom factor, which is related to angle of view:

http://en.wikipedia.org/wiki/Angle_of_view

If you had a telephoto lens with a 10 degree angle of view (equivalent to ~ 205mm lens for 36 mm image), the maximum visible object width at 4500 feet would be 787.398 feet. Your target object width, at 3.5 inches is 0.00037042 or 1/2700 the size of the maximum sized object.

For a still shot camera, something like a Sony A900, 24.6 mega pixel, digital SLR, with 6048x4032 resolution (3:2 apect ratio, it also does 16:9 aspect ratio by cropping this to 6048x3048), with a 200mm lens, would be able to detect the object.

I just realized you meant video camcorder. The high end standard for current display devices is the Sony 4k digital cinema projector, at 4096x2160, cropped to various theater formats. In the past, 35mm or 70mm film was digitally scanned, but there are high end camcorders such as the Red One, that has 4096x2304 resolution if you've got $30,000+ (USA) to spend on a video camera.

http://en.wikipedia.org/wiki/Red_Digital_Cinema_Camera_Company

With a 300mm lens and 24.4 mm image capture width, the angle of view is about 4.6575 degrees. At 4500 feet, the maximum width is 366 feet. The target object is 3.5 inches ~= 1/1255 times the maximum width. The image would occupy about 3 pixels on the camcorder.

For a hi-def camcorder with true 1920 horizontal pixels, for the object to occupy 1 pixel, you need a view angle of 7.12 degrees. Assuming 1x view angle is 45 degrees, then a 10x zoom corresponds to a viewing angle of 4.744 degrees, the object width would be about 1.5 pixels, it would look like a very thin line.

To confirm the numbers, I tested viewing angle versus zoom factor with a Sony HC1 HDV camcorder. Widest angle was about 45 degrees (40 inch width at 48 inch distance), and zoomed at 10x it was a bit more than 4.6 degrees (3 7/8 inch width at 48 inch distance)

Normally, 10x is based on focal length <==> viewing width:

45 ~= 2 tan^-1(39.76/(2*48))
4.744 ~= 2 tan^-1(3.976/(2*48))

On a side note, adding to the spec confusion for camcorders, actual sensor size, is really about 2/3rds of the "stated" size, in this case the "stated" size of the HC1 sensors is 1/3 inch, but this translates into 6.00mm diagonal instead of 1/3 inch = 8.467mm. This should be a spec width = 4.8mm, height = 3.6mm sensor_sizes.htm, but it doesn't correspond to the stated focal length of the zoom lens 5.1mm -> 51mm.

 

[Max CR]

Thanks anyways for your help Jeff, but reds formula worked for me just fine. I was able to calculate it flawlessly. Thank you. I have been looking far and wide for this formula. Thanks. I was finally able to figure it out!

 

[Phrak]

Just curious. What did you get for your measured Lt and Dt?