# How to convert Ohms to luminosity?

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1. Aug 12, 2015

### Nancee_K

So, I recently joined a stargazing club, and for our summer activity, we're supposed to use a telescope, a photoresistor, and a multimeter to measure the luminosity of the moon. I know that luminosity can be calculated with the equation, L = σ AT 4, and that brightness can be measured with the equation b=L / 4πd2. But I don't understand how I could convert convert the data that I recorded in ohms to luminosity. The club president said that we wouldn't need to use physics beyond advanced physics from the high school level, but I don't remember this from my classes. t noticed that some moon sites used illumination percentages (i.e. illumination 54% (what would the value of 100% illumination even be?), or relative brightness percentages (i.e. relative brightness 3%). how would I be able to tie all of these values, equations, and percentage together just by obtaining the brightness of the moon perceived by a photoresistor, measured in ohms. P.S. I know that my location is roughly 393,216.24 km from the moon. I would sincerely appreciate any help at all, for I am completely clueless. Thanks so much in advance!

2. Aug 13, 2015

### DEvens

Whatever device you are using to measure brightness will have to be calibrated in some fashion. That is, you need to know what it reads for a standard brightness. Then you need to know how it varies with changes in that standard brightness. This seems to be what the equations you are using for L are referring to. Without even knowing what device it is, it is impossible to tell you much more. Presumably one of the things in $\sigma AT 4$ is the resistance you are measuring. (I'm guessing there is a typo in there someplace. Should it be $T^4$?) Then the calibration of the device will tell you the other two. Then to get the brightness you need the distance. (Again a typo? Should it be $d^2$?)

It is really difficult to know what "some moon sites" mean when they talk about percentages. Could it be they are referring to the fraction of the moon that is illuminated? That is, what fraction is illuminated by sunlight and what fraction is dark?

3. Aug 17, 2015

### Nancee_K

I'm using a multimeter to measure the light intensity perceived by a photoresistor in ohms. Would the standard brightness the apparent magnitude of the moon? Because I don't know how I would convert that into a standard when I'm measuring the brightness in ohms. Also, yes they're all typos. I forgot to put the ^ to indicate that they are exponents.

4. Aug 17, 2015

### axemaster

The value you record will be specific to your photoresistor. Thus, there is no standard to compare it to. Ideally you would have a calibrating lamp of known brightness to set the scale.

In your case, you will probably have to record the value when the full moon is out, and use that as your basis of measurement.

5. Aug 17, 2015

### Simon Bridge

You need to look up the properties of the photoresistor.
The manufacturer will set the relationship when it is built.

6. Aug 18, 2015

### Robmat

As Simon says, you need the so-called illuminance-resistance characteristics for the photoresistor you're using, which the makers will have. An example is here:
But note
1. Your photoresistor may be somewhat different to this, so don't try to use these curves
2. The curves will, however, generally be power-laws, so small errors in measurement can produce hefty changes in the inferred luminosity
3. The graphs are log-log, so they need to be read carefully

Good luck - fun project !

7. Aug 19, 2015

### Nancee_K

Thanks for the help! The light resistance of the particular model of my photoresistor that I'm using is (10 Lux): 30-50 Kohm. The lux value of a full moon on a clear night
is 0.27–1.0 lux. How do lux, lumens, and ohms all relate? How would I convert ohms to lux? Also, is the light resistance of my photoresistor sensitive enough to detect moonlight through a telescope?

8. Aug 20, 2015

### Robmat

Good questions, and I'm sorry this is turning into a boring "fun" project.

The 10 lux figure is a standard reference point for photoresistors, but unfortunately the corresponding 30-50 kohm resistance you've found doesn't help make the conversion - because (as mentioned in my original post) photoresistors have a NON-linear response. That means we can't use simple proportions, where if 10 lux gives 30 to 50 kohm, 1 lux could be taken to give 3 to 5 kohm, 0.1 lux gives 0.3 to 0.5 kohm etc.

By the sound of it, whoever gave you this project really should have given you much more detail.

9. Aug 20, 2015

### davenn

you would probably need a calibrated variable output light source to do that
then you could graph the results

EDIT: -- or maybe the datasheet for your LDR may already have that graph
google the part number of the LDR

possibly ? ... you haven't given any details on the type/size of scope ?

Dave

10. Aug 20, 2015

### Nancee_K

Oh, right, It's the meade 10" LX200-ACF has 10” diameter f/10 ACF optics (focal length 2500mm) if that helps.

11. Aug 20, 2015

### Nancee_K

Thanks so much for your help! Yeah the person who told us to do this wasn't very descriptive... and my friend got pretty linear values like 12 ohms one night then 10, then 7.2, or something like that. What kind of response do photoresistors even give? I guess I'll keep all of my values in ohms.

12. Aug 20, 2015

### davenn

nice scope I have the Celestron CPC925 (9.25") also f10 and similar focal length ( a little shorter)

That would usually refer to the amount of the moon lit by the sun. 54% would be approx first or last quarter. 100% full moon
you may find elsewhere on the net percentages for other phases eg gibbous phase etc

your readings are probably going to be very different to anyone elses. Unless they use the same scope, eyepiece, photoresistor, multimeter
make sure that the photoresistor is ONLY getting light from through the eyepiece and no ambient light leaking in from the sides from other sources

so produce a graph with the values you measure assoc with difference phases of the moon

you would put moon phase (illum%) along one axis and resistance value along the other axis

Dave

13. Aug 20, 2015

### Nancee_K

That makes so much sense! Thanks so much for your help!

14. Aug 20, 2015

### davenn

you are welcome

come back and post your graph when you get it done

just remember, there is no one correct answer ... all depends on gear being used to take the measurements

D

15. Aug 21, 2015

### Robmat

Wd second that...plus, if you keep all your raw data you will
(a) be able to do the conversion using any further insights you get (eg using the calibrated light source as davenn wisely pointed out, and the makers info)
(b) be able to prove you actually did the hard work, but fell at the last hurdle because of tricky technical issues that I would argue should have been much better explained
(c) do much better than many "real" scientists, who don't keep their data and/or refuse to show it to other researchers - something that's becoming a serious concern here in academia

Upshot - don't beat yourself up, and keep on pluggin' away !

16. Aug 21, 2015

### Nancee_K

This is the data sheet for the type of resistor that i'm using: http://selfbuilt.net/datasheets/GM55.pdf. Does light resistance represent the range at which the resistivity will go to when exposed to light? So if it's 30 - 50 k ohms, then 50 k ohms is the highest, and if the dark resistance is 3 m ohms, then that's the maximum it goes to when it in the dark? if so, I'm pretty sure my data's wrong.