Originally posted by dadudemon
And what I was saying is: it doesn't work out quite like that due to the relationship of cross-sectional area.Same as above: doesn't work out quite like that due to the relationship of cross-sectional area.
You actually get stronger the smaller you get in relation to your own mass (assuming we have an adjustable shrink ray, of course. 😄)
So it would be geometric relationship, not a linear one (we know this, actually).
Not quite for reasons I outlined above.
Ok, let's say a 6 ft tall guy weighs 200 lbs and can lift 200 lbs and he lifts it 2 feet in distance. YAY for English units!!
Shrink him down proportionately to 3 ft tall, and he now weighs 1/8 as much (kind of like your cube diagrams) but he has 1/4 the strength (cross section area goes down 4X when diameter or radius goes down 2X). So his is, in effect, "twice as strong" for his size since he can lift 50 lbs but only weighs 200/8 = 25 lbs. Now he is twice as strong, but remember he is only lifting the weight 1 ft now instead of 2 cuz his arms are shorter. But notice that while he is "stronger" for his size, his is not more "work-ful."
Before shrinking, he was lifting 200 lbs X 2 ft = 400 ft lbs
After "midget-izing" he is lifting 50 lbs X 1 feet = 50 ft lbs
This is difference of a multiple of 8X, which is by how much the total weight decreased. So the ability to work stays in proportion to one's weight assuming that your proportions stay the same with scaling.
Power is work divided by time, and power is what we would want to measure for throwing things I believe. And I think that the power ratios stay related to overall weight just like the work ratios. It would get a little more complicated cuz you could have the bigger guy throwing things a short distance that the smaller cuz could not lift and therefore have a distance of "zero" and not be useful for comparison/measurement.
Does this seem correct or am I pulling it our of my rear?