[datadata]

OK, let's make this a multiple choice question. Assume bitcoin mining machines magically each become twice as powerful, AND twice as heavy, but there are half as many of them, so the total energy and resources used is exactly the same. According to your argument is this change:

a) good, because now twice as many miners actually solve a block, so instead of 99.7% waste it is 99.4% waste.

b) doesn't matter, because I just changed the unit of measurement-- each machine is just doing twice as much work.

c) doesn't matter, because bitcoin is already infinitely wasteful-- in which case pointing out that 99.7% of machines are somehow more wasteful than the other 0.3% does not make sense.

d) bad, for some other reason.

[arcticbull]

Same quantity of power consumption, same quantity of electronics, same waste. The more powerful the machines get the more 'difficult' the 'problem' is to solve. These machines don't do any work unless they successfully guess the nonce. The bulk of its time is spent wasting power. Doesn't matter how many boxes you split it into or combine it into.

[datadata]

Would you then agree that your original point that "99.7% of machines do not solve a block" is not by itself an argument that bitcoin is more or less wasteful than it needs to be? Because through a bookkeeping trick you can reduce or increase that figure by an arbitrary amount.

[arcticbull]

Was that really what this thread was about? It's just a way of visualizing how wasteful it is because most people just don't realize what 50,000 tons of electronics per year thrown out means. Or what 100 trillion watt-hours means.

It's 99.7% of machines based on the characteristics of the current most efficient miner.

I gave all the relevant waste metrics. Power, weight. And to help visualize, the quantity of the current best-in-class miner.

This is legendary pedantry. You can multiply and divide out to get any of the three measures of the scale of waste from the others. This whole thread was about you disliking 'P' in 'P=IV' but having no issue with 'I', 'V' or the idea you can multiply them together.

Gonna go ahead and end our conversation here.

[datadata]

Look, you are touting a metric that you agreed has no bearing on the argument you are trying to make. The number of machines does not matter, it isn't part of any meaningful equation at all. Akin to summing the miles per gallons per car over the number of cars to measure efficiency.