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How does the size/length of an object vary with distance?
Is it a logarithmic relationship? exponential? linear?
I plotted a curve of the size/length of an object for different distances from the camera, and the curve looked exponential/logarithmic. I was trying to understand the reasoning behind that.
Inversely linear is a good approximation.
Imagine a 1,7m tall girl at 1 m distance b. Her head is at point B.
How does the size/length of an object vary with distance?
Let the girl walk away from you. Her size a stays the same. She appears smaller, because she is appearing under a smaller angle. Her angular size changes. Try to imagine it with the picture attached. Using arctangent to calculate her angular size is the correct way. For small angles you can simplify:
Angular size is inversely proportional to its object distance, without using optical devices.
An object on full-field with focal length of 12 mm would be measured incorrectly. An error 2-5% in length measurement may be made. For fish-eye lenses this may be even worse. Hands-on rule: Use the inverse relationship if angular size is smaller than 10°.
The relationship is a simple inverse, i.e.
object size in image = Object size * focal length / object distance from camera
If you keep the same object and the same focal length you get: size = 1/ distance (the =-sign should be proportional-sign).
I'm sure this is a duplicate, but I can't find a good answer to the question in the archives so here goes.
The relationship between object size and distance is an inverse linear relationship, i.e. size is 1 / distance. This makes sense when you think about it as if you double the distance the size halves.
This is why you appear to be observing an exponential: the exponent is -1, if you take the reciprocal of the size, your graph should be a straight line.
It depends on what 'size' means in the question.
Each linear dimension of an object will halve as distance from the camera is doubled and each linear dimension of an object will double as the distance from the camera is halved.
The area of the sensor of film upon which an object is projected will quarter when the distance to the camera is doubled and will quadruple when the distance to the camera is halved.
To put it another way, so long as the subject fits in the frame, doubling the focal length potentially allows recording four times more information with the sensor. Composition aside, that's really the important thing. Thus in terms of focal length:
Because doubling the focal length halves the angular field of view in both dimensions the area of the sensor upon which the object is projected quadruples.
Likewise halving the focal length, quarters the sensor area upon which the object is projected.
In practice this means that going from a 200mm lens to a 300mm lens more than doubles the degree to which a distant subject fills the frame. It's why a 18mm lens is a lot (rather than a little) wider than a 24mm. A 1.4x teleconverter doubles the area the subject projects on the sensor and a 2x teleconverter quadruples it.
The scale of the image can be expressed as a ratio. We are talking about the actual size (height) of an object vs. the height of its formed image. This ratio intertwines object distance (symbol u) with image distance (symbol v) and focal length (symbol F). These are the traditional symbols used in optical formula. I would like to know the origins of these symbols so if anyone knows, please enlighten me.
The focal length of a lens is the measured distance from rear nodal of the lens array to the focused image when the object being imaged is at infinity. This lens-to-image distance elongates if the object is closer than infinity. While the focal length of the lens does not change, this elongated back focus is substituted for the focal length. However, the following equations provide suitable results using F as engraved on the lens barrel for all subject distances except for close-up work (less than 1 yard /meter).
r = v ÷ u (strictly a linear ratio)
Note: use the same unit of measure when solving the equation (I choose millimeters)
Suppose a 1 meter height object is imaged from 10 meters distance using a 50mm lens F = 50mm (focal length) v = object distance = 10m X 1000 = 10,000mm u = image distance (for 50mm lens) 50mm o = object size = 1000mm r = 10000 ÷ 50 = 200 For an object 1000mm height, image height will be 1000 ÷ 200 = 5mm
Suppose a 1 meter height object is imaged from 5 meters distance using a 50mm lens
F = 50mm (focal length ) v = object distance = 5m X 1000 = 5,000mm u = image distance (for 50mm lens) 50mm o = object size = 1000mm r = 5000 ÷ 50 = 100 For an object 1000mm height, image height will be 1000 ÷ 100 = 10mm
When doing close-up work F must be elongated to equal the back focus distance To accomplish we convert focal length and object distance to diopter units.
Suppose a 4mm height object is imaged from 200mm using a 50mm lens Find the revised F or back focus distance 50mm lens express in diopter units 1/50 X 1000 = 20d 200mm object distance expressed in diopter units 1/200 X 1000 = 5d Subtract to find back focus 20 – 5 = 15d Convert back to millimeters to find back focus 1/15 X 1000 = 66.7mm Back Focus
F = 66.7mm (focal length ) v = object distance = 200mm u = image distance (for 50mm lens back focus distance ) 66.7 o = object size = 1000mm r = 200 ÷ 66.7 = 3 For an object 4mm height, image height will be 4 ÷ 3 = 1.3mm
