Alferd Posted December 31, 2022 Posted December 31, 2022 What is the optimum radian and diameter of convex mirror should be apply to achieve the sunlight reflection to cover 5 meters height of wall? Thank you for advance.
Genady Posted December 31, 2022 Posted December 31, 2022 19 minutes ago, Alferd said: What is the optimum radian and diameter of convex mirror should be apply to achieve the sunlight reflection to cover 5 meters height of wall? Thank you for advance. Optimum what?
Lorentz Jr Posted December 31, 2022 Posted December 31, 2022 (edited) How small (short and narrow) does the mirror have to be? If it can be 5 meters (the full height of the wall), it can just be flat (r = infinity) and the wall will be lit evenly. 2 minutes ago, Genady said: Optimum what? Radius? Edited December 31, 2022 by Lorentz Jr
Genady Posted December 31, 2022 Posted December 31, 2022 Radius/diameter can be anything between 0 and infinity.
Lorentz Jr Posted December 31, 2022 Posted December 31, 2022 (edited) Radius/diameter = 0.5! 😁 Edited December 31, 2022 by Lorentz Jr
Alferd Posted December 31, 2022 Author Posted December 31, 2022 Thank you for your reply. 6 minutes ago, Lorentz Jr said: Radius/diameter = 0.5! 😁 😎🙂 9 minutes ago, Genady said: Radius/diameter can be anything between 0 and infinity. How about the optimum radian of convex mirror should be apply if the diameter of mirror is 1 meter width?
Lorentz Jr Posted December 31, 2022 Posted December 31, 2022 11 minutes ago, Alferd said: How about the optimum radian Do you mean radius?
Genady Posted December 31, 2022 Posted December 31, 2022 21 minutes ago, Lorentz Jr said: Radius/diameter = 0.5! 😁 0.5 factorial?
Alferd Posted December 31, 2022 Author Posted December 31, 2022 13 minutes ago, Lorentz Jr said: Do you mean radius? 10 minutes ago, Genady said: 0.5 factorial? I want find the radian, not radius. Radian is a unit for measuring an angle.
Genady Posted December 31, 2022 Posted December 31, 2022 6 minutes ago, Alferd said: I want find the radian, not radius. Radian is a unit for measuring an angle. If I've guessed correctly which angle, then you get the equation 0.5*sin(x)+(0.5-0.5*cos(x))*tan(2x-π/2))=5 OTOH, it can be as small as 0.
iNow Posted December 31, 2022 Posted December 31, 2022 Looks like homework (as I already stated elsewhere when this exact same post was submitted there)
Lorentz Jr Posted December 31, 2022 Posted December 31, 2022 Is this a practical problem, Alferd? Do you really want to do it, or is it just a mathematical exercise?
Alferd Posted December 31, 2022 Author Posted December 31, 2022 26 minutes ago, Lorentz Jr said: Is this a practical problem, Alferd? Do you really want to do it, or is it just a mathematical exercise? This is practical problem.
Genady Posted December 31, 2022 Posted December 31, 2022 3 minutes ago, Alferd said: This is practical problem. I don't believe it.
Alferd Posted December 31, 2022 Author Posted December 31, 2022 12 minutes ago, Genady said: I don't believe it. You means there is impossible to do that?
Lorentz Jr Posted December 31, 2022 Posted December 31, 2022 (edited) 30 minutes ago, Alferd said: This is practical problem. To be perfectly honest, Alferd, if your goal is to light the wall evenly, this is a complicated optimization problem, so I doubt a circular mirror is ideal. Your best bet is probably to buy a bunch of long, skinny, flat mirrors and work out a way to hold them in place next to each other. Then you can adjust their angles experimentally to get whatever distribution of light you want, with no other restrictions on their overall shape. Flat mirrors will also be a lot cheaper than a custom-designed circular mirror. I'm sure people in the Engineering section of this forum will be able to help you. Another way to solve the problem would be to work it out geometrically and iteratively on a piece of paper. Draw a bunch of light beams, reflect them off an arbitrary curved surface, and then adjust the surface and the light beams until the beams have the correct spacing on the wall. Then you can mold your own support, if you know how to do that, and cover it with a reflective sheet. Generally speaking, the mirror will have to be flatter near the top and more curved near the bottom. Edited December 31, 2022 by Lorentz Jr
Genady Posted December 31, 2022 Posted December 31, 2022 6 hours ago, Alferd said: You means there is impossible to do that? No, I mean I don't believe it's a practical problem. Regardless, I gave you the equation above that solves the problem. Is it unsatisfactory? 6 hours ago, Lorentz Jr said: to get whatever distribution of light you want Where in the world the sunlight comes vertically like in the picture? How often does it happen and how long does it last?
Lorentz Jr Posted December 31, 2022 Posted December 31, 2022 (edited) 56 minutes ago, Genady said: Where in the world the sunlight comes vertically like in the picture? Good point! I guess we need the mirrors to be dynamically programmed! Edited December 31, 2022 by Lorentz Jr
Genady Posted December 31, 2022 Posted December 31, 2022 18 minutes ago, Lorentz Jr said: we need the mirrors to be dynamically programmed! Right. OTOH, if it were like in the picture - just as a fantasy - all that would be needed is a flat sheet of aluminum foil curved into a cylinder. Of course, a flat surface like in your earlier post would do, too.
Lorentz Jr Posted December 31, 2022 Posted December 31, 2022 (edited) 24 minutes ago, Genady said: a flat surface like in your earlier post No, my first suggestion was a number of long mirrored strips attached to each other to approximate a curve. Because I've seen those in home-improvement stores. Not strips exactly, but long and narrow. Oh, I see. But that has to be as tall as the wall. Edited December 31, 2022 by Lorentz Jr
Genady Posted December 31, 2022 Posted December 31, 2022 If we assume that the "optimum" means maximum amount of light on the wall, and if we stick with the sketched configuration, then I believe my equation above is the answer.
John Cuthber Posted December 31, 2022 Posted December 31, 2022 Since a flat mirror is cheap and easy and it would work I guess it's optimal.
Alferd Posted January 1, 2023 Author Posted January 1, 2023 23 hours ago, Genady said: If I've guessed correctly which angle, then you get the equation 0.5*sin(x)+(0.5-0.5*cos(x))*tan(2x-π/2))=5 OTOH, it can be as small as 0. From the answer that you given, the curvature is around 55 degree, am I right? 14 hours ago, Genady said: Right. OTOH, if it were like in the picture - just as a fantasy - all that would be needed is a flat sheet of aluminum foil curved into a cylinder. Of course, a flat surface like in your earlier post would do, too. As you said, we will do it in optimum cylinder shape to maximize the light reflect to the wall
Genady Posted January 1, 2023 Posted January 1, 2023 42 minutes ago, Alferd said: From the answer that you given, the curvature is around 55 degree, am I right? I can't say because: 1) according to my calculations, the solution of the equation is about x=87 degrees 2) it is an angle rather than curvature But first you should check the assumptions leading to this equation.
Alferd Posted January 1, 2023 Author Posted January 1, 2023 4 hours ago, Genady said: I can't say because: 1) according to my calculations, the solution of the equation is about x=87 degrees 2) it is an angle rather than curvature But first you should check the assumptions leading to this equation. Thank you for your reply. How about the curvature, do you had any ideas on it?
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now