chr2019 Posted March 5, 2019 Share Posted March 5, 2019 Hello to everyone, Is it possible to write a mathematical formula that will give us the size of an object at various distances? For example, let's say we have a large cube that is 60 feet in size, and we wanted to know how many feet the cube would appear to be when we looked at it at these distances: .5 of a mile 1 mile 2 miles Is there a formula that can accurately calculate what the size of the cube would be when looking at it at these distances? Thank you, Chris Link to comment Share on other sites More sharing options...
mathematic Posted March 5, 2019 Share Posted March 5, 2019 It is easy to get the angular resolution. A=2arctan(30/d) where d is the distance in feet. I am not sure how to convert angle to apparent size. Link to comment Share on other sites More sharing options...
swansont Posted March 6, 2019 Share Posted March 6, 2019 s = r*theta s is the arc length of a circle, r is the distance, theta is the subtended angle The arc length is very close to the linear length (projection of the arc onto a plane), especially if r >> s you can use the angle for the apparent size It's a linear relationship, so if an object is twice is far away, its apparent size drops in half. Link to comment Share on other sites More sharing options...
Sensei Posted March 6, 2019 Share Posted March 6, 2019 15 hours ago, chr2019 said: Is it possible to write a mathematical formula that will give us the size of an object at various distances? You can derive it from Pythagoras formula. Observer is the right side. Object is on the left side. Link to comment Share on other sites More sharing options...
chr2019 Posted March 8, 2019 Author Share Posted March 8, 2019 Thank you for the replies, but I still don't understand how to get the new size for a 60 foot cube when it is viewed at 1 mile away. Link to comment Share on other sites More sharing options...
swansont Posted March 8, 2019 Share Posted March 8, 2019 44 minutes ago, chr2019 said: Thank you for the replies, but I still don't understand how to get the new size for a 60 foot cube when it is viewed at 1 mile away. You haven’t given the distance for the “old” size. Were you an inch away, or ten feet, or 200 feet? If your 60’ tall cube is a tenth of a mile away, it’s the same as a 6’ tall cube 52.8 feet away. Link to comment Share on other sites More sharing options...
chr2019 Posted March 9, 2019 Author Share Posted March 9, 2019 (edited) Is it possible to know the distance of an object when it is at a certain scale? For example, the 60 foot cube at 1:48 scale would be 1.25 feet. If we are looking at a scale model airplane or ship that is in 1:48 scale, what would be the distance? If you are looking at a 1:48 scale model on your desk, and you are pretending that you are actually looking at the real object from a distance, what would that distance be? Edited March 9, 2019 by chr2019 Link to comment Share on other sites More sharing options...
swansont Posted March 9, 2019 Share Posted March 9, 2019 5 hours ago, chr2019 said: Is it possible to know the distance of an object when it is at a certain scale? For example, the 60 foot cube at 1:48 scale would be 1.25 feet. If we are looking at a scale model airplane or ship that is in 1:48 scale, what would be the distance? If you are looking at a 1:48 scale model on your desk, and you are pretending that you are actually looking at the real object from a distance, what would that distance be? You don't seem to be getting that apparent sizes are ratios, because it relates to the angle subtended, while scale is not — it's a linear size. They are not interchangeable. There are multiple pieces of information: distances, the apparent size, and actual size. You can't determine apparent/relative size unless you provide the distance. If you look at a scale model at 1:48 scale, that means 48" converts to 1" in the model. That will always be the case, because you are doing a comparison as if the ruler were right next to the item. It's 1:48 if the model is 1 foot away, and 1:48 if the model is 100 feet away. "How big it looks" requires that you state how far away it is to begin with. A 1:48 model 1 foot away looks the same as the real item 48 feet away. But you have to declare that the model is 1 foot away to come to that determination. Link to comment Share on other sites More sharing options...
mathematic Posted March 9, 2019 Share Posted March 9, 2019 22 hours ago, chr2019 said: Thank you for the replies, but I still don't understand how to get the new size for a 60 foot cube when it is viewed at 1 mile away. Angular resolution makes sense, but I can't see how to define "size". Link to comment Share on other sites More sharing options...
Trurl Posted March 18, 2019 Share Posted March 18, 2019 I never studied optics in physics, but can you guys tell me if a lense would go linear? It sounds like chr2019 is working on a video game. I have a vr headset and distances don't focus in distance to reading writing on the viewpoint. This is why a standard viewpoint using trig and an angle would work for a rifle simulator and eye chart but what do you do if you want to emulate a person's viewpoint? i realize you could always have a linear viewpoint calculated for different perspectives. But I am not skilled enough as a programmer to do this. It seems too memory and computational anyway. There are already Unity and other tools libraries. Isn't this a major problem in optics. My eye doctor said if I could invent glasses that focus near and far I'd be a millionaire. Link to comment Share on other sites More sharing options...
Sensei Posted March 19, 2019 Share Posted March 19, 2019 (edited) 2 hours ago, Trurl said: I never studied optics in physics, but can you guys tell me if a lense would go linear? It sounds like chr2019 is working on a video game. I have a vr headset and distances don't focus in distance to reading writing on the viewpoint. This is why a standard viewpoint using trig and an angle would work for a rifle simulator and eye chart but what do you do if you want to emulate a person's viewpoint? In the case of in real-time VR game simulation, you place two 3D virtual cameras for left and right eye. If person's head is at P 3D vector, left camera is at P-EYE, and right camera is at P+EYE, with EYE being vector with ~3 cm magnitude, direction of vector depends on rotation of the head of player. Majority (if not the all) of currently existing VR headsets can't track human eye movement to change angles at which 3D cameras are pointing at. So focusing is implemented by movement of mouse controlled aim cursor. Game know what 3D object is under mouse cursor, and know what is distance to that object (from analyze of Z-buffer for instance), and can set up 3D camera angles accordingly to have focus on it. Or alternatively it always focus on what is directly in the center of rendered image. It won't work with prerendered static data, like 3D movie, though. 2 hours ago, Trurl said: i realize you could always have a linear viewpoint calculated for different perspectives. But I am not skilled enough as a programmer to do this. It seems too memory and computational anyway. There are already Unity and other tools libraries. Isn't this a major problem in optics. My eye doctor said if I could invent glasses that focus near and far I'd be a millionaire. Astronomers are doing it using liquid highly reflective metal such as Mercury. https://en.wikipedia.org/wiki/Liquid_mirror_telescope Edited March 19, 2019 by Sensei Link to comment Share on other sites More sharing options...
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