A physics problem

[zahc]

Somewhere along the way I forgot how to do math apparently. I can't figure this one out but I'm enjoying puzzling over it. This problem has only nerd appeal, so don't post how physics is stupid.

I am modeling an imaginary physical situation. What I have here in my brain, is a flat, small-area photo sensor in a dark room, pointed squarely at the room's only window, from some distance away. Actually the window is a diffuse, translucent hemispheric dome, 'doming' outward from the room. It is perfectly diffuse, so all parts of it are the same brightness. I need to figure out how close to the window I should hold the sensor, to achieve the maximum reading from the sensor.

I consider the effect on the sensor of an infinitesimal area element of the dome--a point source--so that I can integrate over the whole dome later. The light falling on the sensor from such an element of the dome falls off as a function of the distance from said element (1/r^2), and as a function of the angle between that element and the center of the window/sensor axis (the sensor is a "small area source" in reverse so sensitivity at any nonzero angle theta equals the on-axis sensitivity divided by sin(theta)). I have already done all this, and this is not the problem.

The problem is that I can integrate over the whole dome at any given distance from the window and obtain a "brightness factor" for that distance from the window. Doing so involves integrating over a range of distances becaues of the radius of the hemispherical window, though. Now, what I need is that (now calculable for any distance!) brightness factor as a function of distance from the window. Basically what I need to do is do that integral over the dome surface for every distance from touching the inside of the hemispheric window to infinity. And I don't know how to do that.

In real life what I would do is use a spreadsheet to calculate the integral for a 'lot' of 'closely-spaced' distances, make an XY plot, and pick what looks like the peak. I have already done that. However, I want to be able to do this analytically. But I don't know how to express the value of a definite integral as a function of one of the integrated variables.

[drewtam]

I'll play, but make it easy for me. 

Can you translate the lengthy post into the equations you're using (and want to integrate)?
I assume you have it in spherical coordinates...

Because what you describe sounds like a surface (double) integral over "theta" and "phi" of the brightness or photon flux while holding "r" constant. This assumes you have already derived the equation describing the brightness field. But I'm not sure and want to double check what I think your saying with the actual math.

[bedlamite]

You can't, there is an elephant in the way.

[zahc]

Can you translate the lengthy post into the equations you're using (and want to integrate)?

No, because it's not as simple as that. It's not just a problem of how to solve the equations. I wouldn't be able to solve the problem if the equations were different. And there are no units or numbers...we are only trying to find the distance where the sensor will show the largest reading.

Because what you describe sounds like a surface (double) integral over "theta" and "phi" of the brightness or photon flux while holding "r" constant.

Yes, I think you've got the basic idea. I do indeed want to integrate over the whole inside surface of the dome. In spherical coordinates it would in general be a double integral, but in this case it doesn't need to be a double integral because symmetry around the axis obviates the need to take the one integral. It would just add a factor of 2pi that I can absorb into my master constant. Also, I CANNOT hold R constant, because all points on the dome will not be the same distance away. The only specific case where I could hold R constant would be if the light sensor was at the center of the hemisphere, in the plane of the wall. Since R is not constant, spherical coordinates don't have a great advantage. In my case I collapsed the problem to 2D and then the dome became a half-circle.

I can already do the integral and get an answer for any distance from the wall. What I need to do is write, as a function of the distance, the value of the integral. It's kind of self-referential, and I'm stuck in a loop.

[Brad Johnson]

The answer is... 3

Brad

[Scout26]

The answer is... 3

Brad

No, the answer is 42.

What caliber is the photo sensor ??

[Brad Johnson]

No, the answer is 42.

Yes, but what is the question?

Brad

[HankB]

Actually the window is a diffuse, translucent hemispheric dome, 'doming' outward from the room. It is perfectly diffuse, so all parts of it are the same brightness.

So what you have is actually a hemispherical, uniform Lambertian source, right? If so, then then source can simply be regarded as a flat disc at the window opening.

Since brightness of the source is uniform, then all that matters to the sensor is the angular subtension of the source.

Since the sensor also is flat, then it will only read - at best! - the light coming in at an angle of incidence of 90 degrees or less. So if you bring the sensor closer and closer to the window, you'll see increased readings until you reach the plane of the window, where the reading will be maximized. Continue to move it inwardly into the concave hemisphere . . . and you'll see no further change.

(What you're describing is roughly analogous to an integrating sphere . . . I'm sure Google will give you plenty of hits if you search for that term. Also look for Lambert's law.    )

[zahc]

So what you have is actually a hemispherical, uniform Lambertian source, right? If so, then then source can simply be regarded as a flat disc at the window opening.

Since brightness of the source is uniform, then all that matters to the sensor is the angular subtension of the source.

Since the sensor also is flat, then it will only read - at best! - the light coming in at an angle of incidence of 90 degrees or less. So if you bring the sensor closer and closer to the window, you'll see increased readings until you reach the plane of the window, where the reading will be maximized. Continue to move it inwardly into the concave hemisphere . . . and you'll see no further change.

No no no. You are being practical. I'm not trying to solve the problem. I already solved the problem and built the device that raised the question (using your assumptions which are probably valid). I'm going back, months later now, to fix my mathematical model that I started in my original post...the problem of defining a definite integral as a function of one of the variables it's integrated over.

[drewtam]

I was asking a more basic question... what is your field equation? and what is your surface integral?
Are you taking into account a flat sensor compared to the inverse radius characteristics? or are you ignoring that part and model the equation as if the sensor is domed to match the constant radius?

From what I think I understand, it sounds the like the answer is always an integral over the angles, but never r. So you should be able to do an indefinite integral (0 to theta, 0 to phi). That is what I meant by holding r "constant".
(Which would return a function, F(θ,φ,r), that could be differentiated with respect to r to find the maximum field value.)

But this proposed solution depends on the equations you're using to model the behavior you are looking for (see first paragraph).

[zahc]

what is your field equation?

What field?

Are you taking into account a flat sensor compared to the inverse radius characteristics? or are you ignoring that part and model the equation as if the sensor is domed to match the constant radius?

The sensor is flat. The inside surface of the dome is domed.

it sounds the like the answer is always an integral over the angles, but never r.

r is not constant, and is in the integral. Since part of the equations are 1/r^2. r is only constant for one particular point. I will scan some diagrams tomorrow if I can.

[Physics]

Model it.  Multiple frequencies might make it a little difficult but I could probably bust it out in a Finite Difference Time Domain software in a short period of time.  Of course, running the model might be time consuming, depending on how accurate you want your answer.  

If you send me some schematics, I could probably do it for you in Remcom's XFDTD, but not anytime soon.  I have a talk at the end of the month, working on 3 papers, and trying to defend by mid-May.  After May I could do it though, or maybe sometime before if I can manage to find a bit of spare time.

Of course, there is always the analytic method.  That, I will have to think more about.

How thick is your window?  Is it another variable?

This is my mental picture (side note: I love solidworks):



Sides look like they're at an angle because of the perspective I took.  Quick and dirty, nothing groundbreaking. :D

[zahc]

Of course, there is always the analytic method. Wink  That, I will have to think more about.

I'm ONLY interested in the analytic method. I've already done it with computers. The problem I have with the analysis is I get this far:

definite Integral(some function of sensor-distance) = relative brightness at that point

This integral can be solved at any distance to find the relative brightness at that distance. However I need to write the relative brightness as a function of distance. What I have now is an algorithm for solving the problem at any distance, but that's not good enough.

Your picture is correct. The thickness shouldn't matter. The brightness of the sphere might vary (polka dots, density gradient in the plastic) but said brightness is a known function.

I'm starting to find a hole in my analysis. From the beginning I assume a constant wall-distance and then write the integral. I think I need to keep that a variable and keep it completely separate from the light-distance. Then integrate over that instead of any of the other dimensions in the problem.

 

[Physics]

After thinking about this for a bit, I came to the conclusion that this isn't as bad as you might think.  I would calculate the position and size of your focal volume.  Treat the window as a lens.  Then, after the focal volume, you should see the irradiance drop off as a function of cross sectional area (it should be something like Inaught/Area).  This comes from the fact that the cross sectional area is a function of r.  The further from the focal volume you are, the more "spread out" your light is, and thus the irradiance is lower.

I say irradiance because in optics, intensity is a measure of watts/steradian, or solid angle, while irradiance is watts/m^2, or cross sectional area.  Since your sensor is not domed like your window, I'd use irradiance. 



The above image helps explain the difference to everyone else.  The tip of the triangle is the center of your focus.  The straight line connecting the lower points would be irradiance.  The arc that connects the two would be intensity.  Revolve the image in your head 180 degrees and you will see what I mean by solid angle vs cross sectional area. 

I need to figure out how close to the window I should hold the sensor, to achieve the maximum reading from the sensor.

Calculation of your focal volume will give you this answer. 

[coppertales]

You guys are making my head hurt.  I forgot all that years ago because I did not use it.....chris3

[Brad Johnson]

I still say the answer is 3.

Brad

[Gewehr98]

Ahem.

"42".

[280plus]

OMG! I understand that!   

But what does the symbol up in the angle at the top mean?

[Brad Johnson]

Ahem.

"42".

Brad

[280plus]

Compromise and call it 22.5

[Physics]

Compromise and call it 22.5

Like I tell my students all the time, -1 point for lack of units.

[Brad Johnson]

[Brooklyn accent] I gots ya units right heaah. [/Brooklyn accent]

Brad

[280plus]

Like I tell my students all the time, -1 point for lack of units.

Wha?? I lost my unit?? Uh, ummm, oh, never mind...   

[Regolith]

OMG! I understand that!   

But what does the symbol up in the angle at the top mean?

It's theta, which is simply a variable to represent an angle. 

[280plus]

Thanks, I figured it was a variable angle but didn't know it's proper name.