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"Burning Power" unit of measurement please?

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"Thats why we would use black electrical tape. Its black because it is awesome at absorbing the light we see"

ut you do ignore how awesome is at absorbing the light we don't. And blurays give out a great part of UV.

yours,
albert
 

Meatball

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UV? Its been established here on the forum that 405nm light is not UV. It is a deep violet, but not UV. UV is < 400nm. But regardless of the fact that 405nm light is borderline visibility- wise, I think it's easy to see that 405nm is absorbed by electrical tape since it tends to melt under it.

-Like I said before, we CAN SEE 405nm light, some of us more or less so than others, but in all technicality, 405nm light is visible light.
 

RA_pierce

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You know... this is a good idea, but a power reading and measure of current input is much more meaningful to me than how many layers of electrical tape a laser can cut through.
It would be easier to stick a laser in front of a meter than set up my whole 5th grade science fair project for a silly "power measurement."

If you were to do this, I would recommend setting a fixed scale from 0-10.
A laser with 0"bp" would not burn. A laser with 10bp would set cardboard on fire.
If you start with that you can then set other points within the 1-10 range that mark other burning tasks of increasing difficulty... but the problem is... the chart would start to look like this:


You see what I'm saying?
 

GooeyGus

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Power is power.

100mW is 100mW regardless of wavelength.

Your only MAJOR variable is the material itself that the laser is being applied to. Blu-ray seems like it burns very well because it is readily absorbed by many different materials, including white paper.

Another variable, which is pretty minor in my opinion (because while blu-ray CAN focus slightly tighter than 650, the spot size we're talking about at this point is so minimal regardless of wavelength, that I dont think it makes a huge difference) is how small the spot itself will focus. 100mW focused into a 1mm spot has more energy density than 100mW focused into a 2mm spot.
 
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Power is power.

100mW is 100mW regardless of wavelength.

Your only MAJOR variable is the material itself that the laser is being applied to. Blu-ray seems like it burns very well because it is readily absorbed by many different materials, including white paper.

Another variable, which is pretty minor in my opinion (because while blu-ray CAN focus slightly tighter than 650, the spot size we're talking about at this point is so minimal regardless of wavelength, that I dont think it makes a huge difference) is how small the spot itself will focus. 100mW focused into a 1mm spot has more energy density than 100mW focused into a 2mm spot.
That's what i'm trying to tell him.

Anyway, the phr for example, is not 100% 405nm, there is a lot of invisible in it.
 

Meatball

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Power is power.

100mW is 100mW regardless of wavelength.

Your only MAJOR variable is the material itself that the laser is being applied to. Blu-ray seems like it burns very well because it is readily absorbed by many different materials, including white paper.

Another variable, which is pretty minor in my opinion (because while blu-ray CAN focus slightly tighter than 650, the spot size we're talking about at this point is so minimal regardless of wavelength, that I dont think it makes a huge difference) is how small the spot itself will focus. 100mW focused into a 1mm spot has more energy density than 100mW focused into a 2mm spot.
Correct, power is power. But the differing wavelengths between two beams makes for a difference in the amount of "work" that the power can do.

"Now, not all photons are created equal. The shorter the wavelength the more energy the photon has. Hence a single photon at 405nm is much more energetic than a single photon at 1064nm. You can calculate this using the Planck's relationship E=hv. E=energy, h=6.626E-34 J-s, v is the frequency of the photon calculated from the inverse wavelength relationship v=c/gamma. c=speed of light, gamma = wavelength.

Ok for 405nm. v=300,000km/s / 405nm = 7.407E14 Hz
E = 6.626E-34 J-s * 7.704E14 cycles/s = 4.908E-19 J per photon of 405nm

for 1064nm v=300,000km/s / 1064nm = 2.820E14 Hz
E = 6.626E-34 J-s * 2.820E14 cycles/s = 1.868E-19 J per photon of 1064nm"

-Frothychimp

100mw of 405nm does more "work" than 100mw of 1064nm.

there's more math behind this that I've been looking up, I might be onto something. Plank's relationship incorporates wavelength into the work calculation, which is what we need.

The main obstacle at this point, is not finding the amount of work that a beam is capable of, its determining the absorption spectrum of the particular material we deal with. I say though, that the theoretical capabilities of a beam are what I would be interested in. If we had a theoretical material with a blacked- out absorption spectra, then we could then assume that every beam applied to that material, would then apply all of it's work potential to that surface. Then we could say that this laser, burns better than that other laser, taking both power and wavelength into account.

So technically, we could ditch the whole electrical tape deal, and just work with the numbers, let the math take over. In Frothy's calculation above, we can see just how many more Joules of work one wavelength is over another. This factor can be applied to, and combined with the power output of a laser, to get a particular beam's ability to do work. The unit we can use, is Joules.

Frothy mentioned to me that not necessarily the dot size but the energy density of the beam would have to have standardized. Fortunately, it can be accounted for by standardizing the W/cm^2. So if different dot sizes are used to measure the energy density, proportions can be used compensate for them.

So basically we CAN use math to compare laser beams so as long as we know the laser's wavelength and power output. I'll be doing some more research and try doing some more for real calculations to see what I can come up with. We can thank Frothy for the mathematical assistance.

some thoughts...
 

k1kb0t

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Bp=(nm-mW)/10
sounds good to me
If I use Lazorman's formula as a template here is what I get.

Bp=(1000-(nm-mW))*.1

nm mW Bp
405 100 69.5
405 200 79.5
405 300 89.5

632 100 46.8
632 200 56.8
632 300 66.8


The scale may not be accurate but it is a starting point.


Jon
 
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Meatball

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If I use Lazorman's formula as a template here is what I get.

Bp=(1000-(nm-mW))*.1

nm mW Bp
405 100 69.5
405 200 79.5
405 300 89.5

632 100 46.8
632 200 56.8
632 300 66.8


The scale may not be accurate but it is a starting point.


Jon
I believe the scale is off... From working the comparisons between 1064nm and 405nm using Frothy's method, I get a power difference that says BR does ~2.62 times more work than a 1064nm PER UNIT OF POWER.

Using this method you tried, gives me a factor that says BR has 3.6 times more (BP) than a 1064nm. Double check my calculations in case I missed something major, I only tried it once.

For Lazerman's method, what is the output unit? What is BP measured in?

k1kbot, thanks for experimenting. This is how we'll better get this straightened out. By experimenting, thanks guys!
 

lasersbee

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That's what i'm trying to tell him.

Anyway, the phr for example, is not 100% 405nm, there is a lot of invisible in it.
WHAT... :eek:

As far as I know a non DPSS Laser or Laser Diode puts out a single
wavelength of coherent light.. Correct me if I'm wrong....:undecided:

What other "invisible" is in it.....:thinking:

Jerry
 

HIMNL9

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Probably he's referring to the UV part ..... but technically, it's not a "part", is just that the 405 emission is partially invisible, or at least on the edge of the non-visible UV part, so your perception let you trust that "part" of it is invisible ..... it's a subjective matter :)

About the power scale, what about a scale indicating how much time pass between the moment you point the laser on the "low back" :p of your friend (without say him it), and when he jump out shouting "YOWCH" and starting to curse at you ? ..... (j/k :D)
 

Meatball

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WHAT... :eek:

As far as I know a non DPSS Laser or Laser Diode puts out a single
wavelength of coherent light.. Correct me if I'm wrong....:undecided:

What other "invisible" is in it.....:thinking:

Jerry
I believe you are correct. A blu- ray diode is going to be at least 95% 405nm, and then a little bit of 404nm and 406nm thrown in perhaps. Still nothing to be regarded as "invisible". Most humans can still see light well below 400nm, but the accepted wavelength is ~400nm.

DPSS lasers will have a larger amount of IR or near IR present in the beam. This IR can still however, be translated into heat-energy. Though not usually significant it can still be worked with. I will be in contact with Frothy for some time more, it looks to be for me, that math will be the most accurate way to go.
 
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Just throwing an idea here, but a Sharpie is pretty much universal and absorbs most wavelengths (though is kind of transparent to green? I'm not sure), and so is receipt paper. What about a sharpied dot on a receipt, with a 3 second burst of a laser at the most focused possible. Then the end resut is, the larger the hole burnt in the paper, the more powerful the laser is. Whilst it is by no means accurate, as a rule of thumb this should give a +-50mW "guestimation" of power in the 30-200mW range. This would only be useful for "noobs" trying to diagnose their pointers, though.
 
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Just throwing an idea here, but a Sharpie is pretty much universal and absorbs most wavelengths (though is kind of transparent to green? I'm not sure), and so is receipt paper. ......
Receipt paper is anything but universal. Sorry. I always have at least 4-6 receipts in my wallet from debit card purchases everyday and they are all different UNLESS you get them from the same store like Walmart.

Example, a gas station receipt is nothing like a WalMart receipt or a pizza receipt.
 




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