Recent content by donjoe

Cool, thanks, I wouldn't have thought of the term "albedo" in a non-climate context. :) So from Wiki and from this we have so far: Material, Albedo New asphalt, 0.04 - 0.05 Black acrylic paint, 0.05 Aged asphalt, 0.1 - 0.12 Conifer Forest, 0.08 - 0.15 Bare soil, 0.17 Deciduous trees, 0.15 -...

Does anyone know of any database/list of the reflectivities of various common materials one would be likely to encounter while shining their lasers around? You know, concrete, wood, plastic, clear glass etc. This kind of information could be very useful in deciding what stuff to stay away from...

That was, of course, wrong - pure green at full saturation and half luminance is RGB=0/255/0. The rest was right.

Nah, that's just your imagination running away with you. What I'm doing is avoiding repeating myself while some of you keep asking questions already answered up-thread. I don't like being repetitive or pretending not to understand plain English text, like you, so I prefer to move on to cracking...

Because if he buys better sunglasses he will also have the benefit of, well, better sunglasses. You know, for going out in the sun. :D

Then you need better sunglasses and/or better eyes. :P

I've already discussed the worst-case-scenario aspect of those regulations. Except when you aren't. Margins are good, but different people can still disagree on what the right margin is. When you think about it, the safest thing to do is to just never use lasers at all (and maybe walk around...

Unless they have a friend with a power meter. :P But for 80% absorption or less you may not even need a power meter - this kind of transparency can be checked with much easier and cheaper methods, e.g.: - Display an area of pure green somewhere on your monitor (pure green means RGB=255/255/255...

No indeed. What I said was "typical troll answer", with emphasis on "answer". A single troll-like remark doesn't make you a troll generally. So no, I didn't actually call you a troll. :) That's because you're not paying attention. My question was not the same as the one that's been asked around...

Cheaper, like I said. Typical troll answer. If you don't know what physics is or what math is or don't care about arguing for your assertions (instead of throwing them around like rocks), by all means feel free to ignore this topic. Let's be serious here, that list isn't really that long. :) The...

"Pretty well"? :) Even with glasses that absorb 80% of incident light, it still means you will see 1mW of your 5mW passing through, which will produce a clearly visible hotspot even in daylight, as many class II laser pointers do. But making the laser completely invisible was never the point for...

But then as I explained, there isn't "more light" available to do any extra damage, since (using your example) we're still talking about 25 mW being reduced to 5 mW by an 80% absorptance. Pupil dilation doesn't matter, you're still only getting 5 mW into your eye. You'll blink and you'll be...

In fact I don't think it is all they do. In addition to making your pupils dilate (which I have explained above is not a concern when you're talking about a constant safe level of incoming power), they also absorb 33% to 90% of visible light, which could very well include the radiation of your...

I don't suppose you care to give any explanations with that.

As I understand, sunglasses can have an absorptance anywhere between 3:1 and 10:1, so I'm thinking if this is evenly distributed across the spectrum you should be able to get adequate protection with sunglasses while using lasers with powers between 3 x 5mW and 10 x 5mW (depending on the quality...