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Random request: Help calculate the minimum power density needed to pop a black balloon

CurtisOliver

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As the thread title suggests,
I am asking for those with a bit of spare time on their hands to help me out with a fun little problem.
I'm not going to give away the reason why yet, but I appreciate it is a random request.
These are the conditions:
  1. The laser/s used must be LPM'd
  2. The laser must not be focused in any way
  3. The dot size must be accurately determined on both axes.
  4. The balloon must be black.
I will then do the calculations myself.
Thank you. :)

It's probably best that you avoid testing with high power Class 3b's/4's. Stick to <100mW.
 



CurtisOliver

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I’d say maximum 3-4 seconds. Doesn’t necessarily have to be instant. I’ve basically got to simulate a much more powerful laser over a large distance.
 

paul1598419

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Are you looking for the least power that will pop a balloon? I don't see what you are trying to do here.
 

Anthony P

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I have a CO2 laser that I can vary output power and meter. Also expand beam and measure diameter to calc power density. Just need to find some black balloons. Let me know if you think this could help.
 

CurtisOliver

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Are you looking for the least power that will pop a balloon? I don't see what you are trying to do here.
Paul, specifically I am looking for the least power density to pop a black ballon. Because power alone isn’t enough to know on it’s own. If I can work out the rough W/cm2 required, then I can calculate how far away I can pop the balloon with a Class 4 with known beam specs.

I have a CO2 laser that I can vary output power and meter. Also expand beam and measure diameter to calc power density. Just need to find some black balloons. Let me know if you think this could help.
Shame it’s a CO2 as your setup would of been perfect. Ideally it has to be a visible wavelength as the absorption will differ at that part of the spectrum. I’m not sure how much the absorption changes with something like a black balloon however. Probably not much that can be found out with a trivial object like that.
Thanks a lot anyway.
 




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