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Hi I am newbie here and this is my first question so please do not mind if I don't use the correct language or anything.

I want to know the relation between size of a concave mirror ( measured by radius of curvature I think ) and amount of heat produced at its focus when concave mirror is kept in sunlight. If climatic conditions matter ( I think they do ) then assume them to be similar to coastal regions and area should lie in subtropics or tropics. Please provide the explanation for relation.

My question is different from "What temperature is achieved in focus point by 5000 flat 1x1cm mirrors onto a satellite dish? " as I want to take climatic conditions and latitude into account.

I can ignore climatic conditions if their effect is not more than +-10W on heat produced. You can consider the size to be about 1.5 times of r5800

Existent
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    I believe you will find the answer to your question in my earlier answer. Let us know if that does not sufficiently address your question. – Floris Dec 21 '15 at 14:41
  • Note there are two parameters that define the "size" of a concave mirror: the radius of curvature, and the diameter. As I explain in my other answer, a larger diameter and smaller radius of curvature both increase the potential temperature at the focal point. If their product is constant, then the temperature achieved is (approximately) constant. – Floris Dec 21 '15 at 14:47
  • How much heat in joules will it produce at focus ? – Existent Dec 22 '15 at 14:34
  • Wouldn't climatic conditions matter ? – Existent Dec 22 '15 at 14:36
  • Sunlight is about 1 kW per square meter. Multiply by area to get power (Joules per second). If your mirror is not pointed straight at the sun you have to use the "effective" area (area projected onto the direction of the sun = area times cosine of angle) – Floris Dec 22 '15 at 14:37
  • Could you please elaborate effective area concept. How we would determine the angle? Why wouldn't climatic conditions matter. And location with respect to equator should matter – Existent Dec 22 '15 at 14:40
  • Of course the 1kW number refers to unobstructed sunlight and atmospheric conditions matter. There is no good way of determining the effect of latitude and weather other than measuring it. For angle imagine at what angle the mirror looks "biggest" (that is 90 degrees from where it looks "smallest") then take the angle between that direction and the direction of the sun. – Floris Dec 22 '15 at 15:25
  • Note that the +- 10 W number you quote is relevant only when you specify the size of the mirror. For a 10 cm mirror, where the total power is at most 10W, you can state that the actual number is always within 10W of the calculated value – Floris Dec 22 '15 at 15:28
  • You can consider the size to be about 1.5 times of r5800 – Existent Dec 22 '15 at 15:41

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