Posted: 6/30/2009 9:59:41 PM EDT
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Ok, so I was wondering if someone could help me with this problem.
A light source is located over the center of a circular table of diameter 4 feet. Find the height h of the light source such that the illumination I at the perimeter of the table is maximum if I= k(sin(alpha))/(s^2), where s is the slant height, alpha is the angle at which the light strikes the table, and k is a constant. |
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well im assuming you are an engineering or physics student. I personally would make a computer program or even using excel and changing my variables s and alpha until illumination is a maximum.
I would start out with making alpha some reasonable arbitrary angle and then setting up a plot/graph "I vs s." So the maximum on the curve should be your optimal height for that given angle. Going further, you could try again both increasing and decreasing alpha and seeing if illumination increases (or improves). Essentially I would solve it using an iterative process. If you are an engineering student, I would use matlab, fortran, C, or excel. It would be pretty easy i think if you follow along with what i mentioned. |
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Since k is a constant, it does not matter what it is. Assume it is 1 to keep things simple.
sin(alpha) = h/s (draw a picture) The equation then simplifies to: I = k(sin(alpha))/(s^2) I = 1(h/s)/(s^2) I = s/(s^2 * h) I = 1/s*h The rest of the math is yours. Assuming you gave us the correct equation, the lower the light the greater the intensity. When h––> ZERO, the intensity approaches infinity. Please check the equation. |
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Quoted:
Since k is a constant, it does not matter what it is. Assume it is 1 to keep things simple. sin(alpha) = h/s (draw a picture) The equation then simplifies to: I = k(sin(alpha))/(s^2) I = 1(h/s)/(s^2) I = s/(s^2 * h) I = 1/s*h You got it from there? FAIL :-) i = k(h/s)/s^2 i = kh/s^3 so what use is this? s depends on h. |
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Quoted:
Quoted:
Since k is a constant, it does not matter what it is. Assume it is 1 to keep things simple. sin(alpha) = h/s (draw a picture) The equation then simplifies to: I = k(sin(alpha))/(s^2) I = 1(h/s)/(s^2) I = s/(s^2 * h) I = 1/s*h You got it from there? FAIL :-) i = k(h/s)/s^2 i = kh/s^3 so what use is this? s depends on h. i = kh/s^3 correct s = sqrt(h^2 + r^2) where r = diameter of table = 2 feet. I = k h/(h^2 + 4)^3 Maxmimize h = take derivative and set = 0 d/dh of h(h^2 + 4)^-3/2 = (h^2 +4)^-3/2 +h(-3/2)(2h)(h^2 + 4)^-5/2 = 0 simplify into a fraction then choose the numerator = 0 = 4 - 2h^2 2h^2 = 4 h^2 = 2 h = sqrt(2) ~ 1.4 ft |
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D'oh!! You are right. I'm rusty and made a basic math error.
FAIL :-) i = k(h/s)/s^2 i = kh/s^3 so what use is this? s depends on h. To answer your question - Plot the equation and you'll find a max value of I. If this were differential (beginning) calc, take the derivative of the equation and set it to zero. That will be a max or a min of the graph. The nature of this problem pretty much assures us that it will be a max. |
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Since k is a constant, it does not matter what it is. Assume it is 1 to keep things simple. sin(alpha) = h/s (draw a picture) The equation then simplifies to: I = k(sin(alpha))/(s^2) I = 1(h/s)/(s^2) I = s/(s^2 * h) I = 1/s*h You got it from there? FAIL :-) i = k(h/s)/s^2 i = kh/s^3 so what use is this? s depends on h. i = kh/s^3 correct s = sqrt(h^2 + r^2) where r = diameter of table = 2 feet. I = k h/(h^2 + 4)^3 i = kh/(h^2+4)^(3/2) Now, to find the maximum I, fire up Mathematica and take the first derivative of i with respect to h. Solve for 0's |
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well im assuming you are an engineering or physics student. I personally would make a computer program or even using excel and changing my variables s and alpha until illumination is a maximum.
I would start out with making alpha some reasonable arbitrary angle and then setting up a plot/graph "I vs s." So the maximum on the curve should be your optimal height for that given angle. Going further, you could try again both increasing and decreasing alpha and seeing if illumination increases (or improves). Essentially I would solve it using an iterative process. If you are an engineering student, I would use matlab, fortran, C, or excel. It would be pretty easy i think if you follow along with what i mentioned. adding to what i said above, and after rereading the problem, i have one question to ask, is k a constant that is given or what? Im not sure what exactly k is for in the problem. I am also seeing the problem as a 2 part problem that involves trial and error. If i am understanding what s represents, it is the distance from the light source itself to the perimeter of the table (in other words the hypotenuse of "triangle"). If so i would use trig and find a s value and angle that satisfies the condition for the adjacent side of the triangle, which would be the radius of the table. After doing that, then plug in your s and alpha angles to your original equations and see what your illumin. is. Then try again with a smaller and larger alpha and see whether I increases or decreases based on s increasing. In other words if you increased alpha and found I to be higher than your first trial, then repeat with an angle higher then your previous alpha. You should eventually converge to an optimal I for a given I and alpha, which using trig will give you h. One note on the angle alpha, is im not sure if the angle is from the vertical to the "light" or from the table to the "light", if that makes any sense. |
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Quoted:
well im assuming you are an engineering or physics student. I personally would make a computer program or even using excel and changing my variables s and alpha until illumination is a maximum.
I would start out with making alpha some reasonable arbitrary angle and then setting up a plot/graph "I vs s." So the maximum on the curve should be your optimal height for that given angle. Going further, you could try again both increasing and decreasing alpha and seeing if illumination increases (or improves). Essentially I would solve it using an iterative process. If you are an engineering student, I would use matlab, fortran, C, or excel. It would be pretty easy i think if you follow along with what i mentioned. adding to what i said above, and after rereading the problem, i have one question to ask, is k a constant that is given or what? Im not sure what exactly k is for in the problem. I am also seeing the problem as a 2 part problem that involves trial and error. If i am understanding what s represents, it is the distance from the light source itself to the perimeter of the table (in other words the hypotenuse of "triangle"). If so i would use trig and find a s value and angle that satisfies the condition for the adjacent side of the triangle, which would be the radius of the table. After doing that, then plug in your s and alpha angles to your original equations and see what your illumin. is. Then try again with a smaller and larger alpha and see whether I increases or decreases based on s increasing. In other words if you increased alpha and found I to be higher than your first trial, then repeat with an angle higher then your previous alpha. You should eventually converge to an optimal I for a given I and alpha, which using trig will give you h. One note on the angle alpha, is im not sure if the angle is from the vertical to the "light" or from the table to the "light", if that makes any sense. isotropic radiation is spherical I = power/4 pi r^2 (power spread over surface area of sphere) Here r is slant range s the constant k absorbs the power/4pi terms into one for easier manipulation. |
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