confused about multiple flash output power

I'm a bit confused about using multiple flash units @ reduced power output. If I use a flash unit @ 1/64th power to freeze a subject's movement, what would the output power be if I used 2 flash units @ 1/64th power? Would it remain @ 1/64th? or would it be equal to 1/32nd power? Or perhaps even 1/128th power?

The short answer is you will have two strobes at 1/64 power. Photographers typically use f stops to measure light intensity. Depending on where you place your strobes you may or may not affect the overall exposure of the image when you add the second light.

If the two units are side by side, triggered at the same time, then you will have 1/32 power, but at the 1/64 flash duration. This is one way to get a bit more power when using a short duration to stop action.

This technical stuff can get complicated, what with the inverse square law and all that, but try not to make things more complicated than they need to be. You're over-thinking this.

The basic math is: 1/64 + 1/64 = 2/64 = 1/32.

You double the number of lights (with the same output), you double the light. In photo terms, that's one stop.

Curtis mentioned the inverse square law which is an important part. In your exercise the flashes would need to be the same distance to the subject.

Thanks Doidinho, Phototeacher, Curtis N & gonzogolf for your replies. I had forgotten about the inverse square law. This clears up my confusion. Thanks

Curtis N wrote in post #14666171
You double the number of lights (with the same output), you double the light. In photo terms, that's one stop.

Yes and no, if the lights are side by side and aimed to the same location of the subject yes that will double the light falling upon the subject, however the further the lights are spaced from each other the multiplier of the light drops from just about 2 and becomes lower and lower until the lights are 180 degrees apart, at which point the multiplier becomes for all practical purposes a factor of 1.

Wayne