Size and distance

[70116]

I have been unable to find a formula which allows me to calculate what size to make an object so that it appears to be the size of the object would be at a different distance. That’s somewhat confusing so here is an examples.

My base object is an 8 inch circle.

If I draw an 8" circle and put it 15 feet away, how large a circle would this appear to be? If I put it at 21 feet away? 30 feet away?

What is the formula to perform this calculation?

Thanks for your help.

 

[Loren]

To my knowledge, there is no such formula. When drawing an item, it is done in respect to other items in the picture. For example, suppose you take a picture of a barn and print it on a 4" X 6" piece of photo paper. Now, print the same picture on a 12" by 18" piece of paper. Obviously the barn is the same distance from the viewer (camera?) but the barn door will be of different dimensions.

 

[mmm4444bot]

I'm thinking that your example needs to be more specific.

For example, I put an 8-inch (diameter) ball on a table, and then carried a ruler to a distance 15 feet away.

When I hold the ruler in front of my face, the ball appears to measure just over an inch.

But, this height really increases or decreases as a function of how far from my face I hold the ruler!

How do we measure the perceived height of the ball? I mean, how far should I hold the ruler from my face, when I measure the perceived height?

 

[70116]

I agree with your analysis that actual visual reference and measurement is not a viable option. That’s why I’m looking for a mathematical solution.
Here is what I’m trying to accomplish.

I have taken up the sport of target shooting. One of the targets is an 8” circle. Multiple physical targets with 8” circles are placed at varying distances from the shooting position. These distances are 5, 7, 10, 15 and 25 yards. I am required to shoot at each of the targets within a time frame. It would typically be 2 shots at the 5 yard target, 2 shots at the 10 yard target and 2 shots at the 25 year target within 6 seconds. (The combination of distances can change.)

I would like to practice at a target shooting range which is close to my home but it is an indoor range that is not even 10 yards long and they don’t have the space for me to use multiple targets.

What I would like to do is to create targets that simulate the visual appearance of an 8” circle at the varying distances. I can then place any combination on a single piece of cardboard and practice at the range with space limitations. I will place all of the targets at 5 yards so I can use an actual 8” circle for this target. I’m trying to derive what size circles I need to provide the proper visual references for the other targets.

 

[wjm11]

I have taken up the sport of target shooting. One of the targets is an 8” circle. Multiple physical targets with 8” circles are placed at varying distances from the shooting position. These distances are 5, 7, 10, 15 and 25 yards. I am required to shoot at each of the targets within a time frame. It would typically be 2 shots at the 5 yard target, 2 shots at the 10 yard target and 2 shots at the 25 year target within 6 seconds. (The combination of distances can change.)

I would like to practice at a target shooting range which is close to my home but it is an indoor range that is not even 10 yards long and they don’t have the space for me to use multiple targets.

What I would like to do is to create targets that simulate the visual appearance of an 8” circle at the varying distances. I can then place any combination on a single piece of cardboard and practice at the range with space limitations. I will place all of the targets at 5 yards so I can use an actual 8” circle for this target. I’m trying to derive what size circles I need to provide the proper visual references for the other targets.

That clarifies the situation. Here is an easy solution: multiply the 8" target by the ratio of actual distance/"pretend distance". Example for the 7 yd target:

(5/7)(8") = 5.71" diam.

10 yd: (5/10)(8") = 4" diam.

15 yd: (5/15)(8") = 2.67" diam.

25 yd: (5/25)(8") = 1.6" diam.

 

[mmm4444bot]

I agree with your analysis that actual visual reference and measurement is not a viable option. That’s why I’m looking for a mathematical solution.

Oh, I wasn't intending to conduct experiements! Most mathematical solutions begin with a sketch (at least, in applied scenarios, like yours). But, if I can't get a handle on a scenario as described, then I can't make a sketch in which formulas and relationships between quantities begin to emerge.

?

That clarifies the situation.

Heh, heh. I was confusing myself, before I got a ball and ruler, by drawing various similar circular-based cones, wondering how to come up with an angle. :lol:

 

[70116]

Thank you all for you help.

I was overcomplicating the solution methodology in my thought process, and had gone further astray while trying to find a solution by searching the internet. Adding to this is the 40+ years since my last math class and the lack of any practical application similar to this in the intervening years. As always, explaining the problem to be solved in 'real' terms rather than an attempt at a hypothetical explanation allowed the 'experts' to derive a simple, and elegant solutgion that even I can use to deal with some other scenarios.

Again, thank you for your help.