Question about thermal headed meters

[Laser Ben]

No, it trys to give you a reading in W/Cm^2 but it fails because it does not know the size of the spot. I believe there are some sensors where you can put a specific pinhole in front of the detector, and then it measures the light that gets through it.

[daedal]

No, it trys to give you a reading in W/Cm^2 but it fails because it does not know the size of the spot. I believe there are some sensors where you can put a specific pinhole in front of the detector, and then it measures the light that gets through it.

That would then assumes a 100% uniform beam profile, doesn't it?

--DDL

[Lasernerd]

No, it trys to give you a reading in W/Cm^2 but it fails because it does not know the size of the spot. I believe there are some sensors where you can put a specific pinhole in front of the detector, and then it measures the light that gets through it.

hmm,interesting.
it gives a good solid reading,it knows it's sensor size etc.
I am very aware there is mega math involved on all sides

[mccarrot]

Does someone know the calculations to calcutate the mW/cm2

for example the spot size on the detector is 4mm and de power reading = 100mW.

[Laser Ben]

Yeah, you find the area of the spot, then you devide the power by that, then multiply it by the area you want. Watch the units though, square mm=square cm/100

[Xytrell]

Does someone know the calculations to calcutate the mW/cm2

for example the spot size on the detector is 4mm and de power reading = 100mW.

Like Mr. Ben said, it's a simple geometry problem. you have 100mW in a 4mm beam. You just need to convert diameter to area (it's a circular spot, right ? )

A= πr²

=π(2mm)²

=4πmm²

≈12.56mm²

If you want /cm² instead:

12.56mm² × 1cm/10mm × 1cm/10mm

= 0.1256cm²

So you have an average power density of 100mW/0.1256cm² or simplified: 796mW/cm² If you need a formula:

I = p/(π[d/20]²)

Where I is the average intensity in mW/cm², p is power, and d is the diameter of a round spot in mm.

[mccarrot]

Thank you!!!

The new oproximation to caluclate MPE values with a static beam:

link to original document: http://www.laserist.org/files/audien...rview_2pt1.pdf

A “Level 1” show does not exceed the MPE limits. A quick and easy-to-measure approximation
is to set each laser projector so it does not exceed 10 mW/cm2 in the audience area, when
measured as a static beam at maximum show power, at the point of closest audience access.

(or 100mw/cm2 for the new 10xMPE guideline as 1x MPE not acceptable in europe becuase the show looks dim)

So when I make a 4mm pinhole in front of my detector and do the math I can now calculate and estimate the MPE values.

so with a 4 mm pinhole and 12,56mW reading on the meter it should be 100mW/cm2 = 10MPE

right?

[mixedgas]

[QUOTE=mccarrot;98767]Thank you!!!

The new oproximation to caluclate MPE values with a static beam:

link to original document: http://www.laserist.org/files/audien...rview_2pt1.pdf

Some more help

https://home.comcast.net/~trwalters0...20Analysis.pdf

Steve

[mccarrot]

Pfff thanks, but I wil try the simplified method first.