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FrozenGate by Avery

bs detector lpm experiment

D

drspastic

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ok, starting this thread as a journal.

first test is a block of styrofoam housing a 0.5 gram aluminium sheet blackened with marker pen.
an aperture at the front and back allows laser to enter and hit the sheet, and a infrared thermometer secured to the rear to record temperature change over a given time period.

after that, math is our friend.
 





first pics
 

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Do you know the math behind it?

I was doing some research on the math and came across this very nice calculator. You'll need to cross reference with a table of air properties that can be found at engineeringtoolbox.com or something.

Just bear in mind the relationship between temperature and power is non-linear.
 
first exposure: lazerpoint green 200mw (advertised power)
start temperature 18.6c
1 minute exposure
end temperature 19.8c
1.2c rise in 0.5g mass of Al specific heat 0.91 Kj/Kg/c or 0.22 Kc/Kg/c

maths is not being my friend... it seems 60 seconds gives me 0.546 j
or 9.1 mW!

lets try with a bigger laser....
 
Tw15t3r said:
Do you know the math behind it?

I was doing some research on the math and came across this very nice calculator. You'll need to cross reference with a table of air properties that can be found at engineeringtoolbox.com or something.

Just bear in mind the relationship between temperature and power is non-linear.
Click to expand...

my math skills have rusted like a 1970's fiat.
thats why im posting the readings rather than my dodgy working out.
volunteers are more than welcome to put me straight....please!
i dont think my first working was so good, 9mw????

i used this: http://www.endmemo.com/physics/specificheat.php
 
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cokoo 1000mW blue:

start 19.5c
1 minute
end 59.5

300mW?
 
Last edited:
Ok...

Well, there are a number of factors why your calculation ended up with a power that is that low. What you have calculated is the amount of energy stored in the aluminium piece. It is indeed 9mW, but that's not what your laser is outputting.

The aluminium piece is losing heat to the environment both through radiation and convection, and so that needs to be taken into account, and that's where most of the power has gone. Also, not all of the energy from the laser is absorbed by the aluminium piece. No doubt, you coloured it black and that helps with absorption, but it's not going to be perfect, so there'll be some losses. (There'll always be some losses as long as you can see the dot on the black surface. A perfect black surface won't even allow you to see the dot when you shine on it)

The problem is that, if you use a limited time period, then you'll have to perform a rather tricky integral to get the amount of energy lost in total, and then divide it by time to know the power.

Instead, there's an easier method. If you could shine it on to the aluminium piece until the temperature stabilises, then the amount of power in and power out will be equal. From there, at that temperature, I can use the online calculator to calculate the power loss of your aluminium piece, and making a number of small assumptions, we could get a closer figure to the power of your laser.

Do you think your laser can withstand such a long duration of continuous power? You could also try a less powerful laser to attempt that. I only need the temperature to stabilise, and then these figures from you:
1. The surface area of the aluminium piece facing the laser.
2. The surface area of the aluminium piece facing your IR thermometer.
3. The ambient temperature.
 
Tw15t3r said:
Ok...

Well, there are a number of factors why your calculation ended up with a power that is that low. What you have calculated is the amount of energy stored in the aluminium piece. It is indeed 9mW, but that's not what your laser is outputting.

The aluminium piece is losing heat to the environment both through radiation and convection, and so that needs to be taken into account, and that's where most of the power has gone. Also, not all of the energy from the laser is absorbed by the aluminium piece. No doubt, you coloured it black and that helps with absorption, but it's not going to be perfect, so there'll be some losses. (There'll always be some losses as long as you can see the dot on the black surface. A perfect black surface won't even allow you to see the dot when you shine on it)

The problem is that, if you use a limited time period, then you'll have to perform a rather tricky integral to get the amount of energy lost in total, and then divide it by time to know the power.

Instead, there's an easier method. If you could shine it on to the aluminium piece until the temperature stabilises, then the amount of power in and power out will be equal. From there, at that temperature, I can use the online calculator to calculate the power loss of your aluminium piece, and making a number of small assumptions, we could get a closer figure to the power of your laser.

Do you think your laser can withstand such a long duration of continuous power? You could also try a less powerful laser to attempt that. I only need the temperature to stabilise, and then these figures from you:
1. The surface area of the aluminium piece facing the laser.
2. The surface area of the aluminium piece facing your IR thermometer.
3. The ambient temperature.
Click to expand...

your the kind of friend i need! i also am not sure i should have not used kelvin in the first calc... i shall take the dogs out and retry you scheme.
 
No problem! It's getting a little late here already, and I'll probably be going to sleep soon.
Just post your figures up, and I'll tackle them in the morning. Other members might pop in to help as well!
 
16.6 ambient, 78 stable. 1000mW
17.7 stable from the 200mW
the metal is 20x20mm with 10mm dia port either side for input and measurement. the test piece is held in a 30x40 expanded polystyrene block 20mm thick ie 10mm front and 10mm rear.
the measurement port is sealed to the ir thermometer so no convection airflow to speak of. laser port is recessed by 10mm, the laser head blocks this from the air also.
i think the poly block may make the calculations difficult. i will try again on a bare ali sheet. batteries dead now, that may have knocked it out too! really crap ultrafires. 1800mAh my arse they are! lucky if that give 180

there is definately a simple lab setup to measure lasers without dedicated machines. this may need some more input from thermodynamics experts.
 
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Hmm, yea. The shielded piece of aluminium is probably throwing my calculations off. I got 73.7mW for your 200mW laser, and 691mW for the 1W laser. The factors are off by about 2-3 for the 200mW, and 1.5 for the 1W. It's not too bad to start, but I'm making some bad assumptions here....

If possible, use a piece very thin aluminium (maybe foil). Make it square/rectangle, and don't house it. Colour every part of it as black as you can. That would make the method I'm using most accurate.
 
ok i will try cooking foil later, maybe i have some firepaint to coat a test area. i suppose its best to spread the beam to cover as large an area as possible rather than a sharp focus.
i also changed the batteries in the green, noticed a big difference.
once we get consistent results maybe an lpm owner can try to reproduce the experiment, or at least the owner of a known power laser.
the non contact infrared thermometers are no doubt the key to a cheap diy lpm. light detectors off the shelf (diode, transistor, LDR, solar cell) dont have a linear response, and thermopiles, while coming down in price, still need calibrating to a scale. with the thermometer we started with a calibrated instrument, and disposable cheap if we dont need lab quality:

free shipping LCD Non Contact IR Infrared Digital Pen Thermometer Red-in Temperature Instruments from Industry & Business on Aliexpress.com
(example, not endorsed or tested by myself)

im sure this one is about as crap as the one im using, but mine is good enough to check soldering iron temperature, the oven in my wood stove, and the inside of my freezer. its always a few degrees out but thats a trait of these devices; depends what and how you point them at.
mine is a n19fr from a maplin shop in uk, cost 9 gbp so i just bought it to play with, or scrap for the sensor. its been good for 6 years.
 
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Well, yea, like I said, it's been done here: Simple Laser Power Meter using IR Thermometer
It's just unfortunate that he isn't around anymore to help explain the theory and further develop on it. Also, he works in Imperial Units... :scowl:
Glad you are able to help us (or me, since no one else has replied yet) with this experiment which will both act as a second test to confirm the results of Warske who did the previous one, and to also hopefully develop this measurement technique further.

I would like to see how far does a linear approximation hold actually, because according to the math I've done, temperature rise in steady state (after stabilisation) is not linear with the power applied. And I have a hypothesis that the inaccuracies that people are facing with LPMs at high power causing under-measurements has so far not been proven or disproven. (more about it here: Power Losses in Thermal Measurements of Lasers (Theoretical))
 
15.625mg foil (10cmx10cm weighed @2.5g then divided into 16ths)
625 mm2 square coloured black.

does that sound a feasible test piece?
 
Tw15t3r said:
Well, yea, like I said, it's been done here: Simple Laser Power Meter using IR Thermometer
It's just unfortunate that he isn't around anymore to help explain the theory and further develop on it. Also, he works in Imperial Units... :scowl:
Glad you are able to help us (or me, since no one else has replied yet) with this experiment which will both act as a second test to confirm the results of Warske who did the previous one, and to also hopefully develop this measurement technique further.

I would like to see how far does a linear approximation hold actually, because according to the math I've done, temperature rise in steady state (after stabilisation) is not linear with the power applied. And I have a hypothesis that the inaccuracies that people are facing with LPMs at high power causing under-measurements has so far not been proven or disproven. (more about it here: Power Losses in Thermal Measurements of Lasers (Theoretical))
Click to expand...

is this where the 'specific heat' of the test material comes into play?
 
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