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Will_Sawin comments on Inverse Speed - Less Wrong
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I can't quite tell if your comment is analogous to "move your arms and legs!" or if instead you're just agreeing with what I said here (point 2), and mistaking the nature of my difficulty.
In any case, "wanting people to explain things to me intuitively" is not at all how I would put it. In my experience, asking for an "intuitive" explanation will often produce the very opposite kind of explanation from the kind I want: concrete and ad-hoc instead of abstract and general.
If I can't figure something out, it's because there's a concept that I'm missing, and what I want is to be told about that concept. The concept doesn't have to be "intuitive", it just has to be generalizable and (preferably) concisely expressible.
However, most people's approach to explanation is antithetical to this. They try their best to avoid introducing new concepts, tortuously beating around the bush as a result. It's not that I can't follow their explanations, but rather that I can't properly generalize from them, because they're not telling me the "real story".
In terms of your metaphor: my difficulties arise from lacking a hammer, not from being unwilling to smash.
My understanding has been that if you know how to set up a problem as a formula then you know how to make it out of toothpicks and rubber bands so you understand it.
This requires a certain ability to manipulate formulas:
2xy/(x+y)=2/(1/x+1/y)=1/(((1/x)+(1/y))/2)
as I'm sure you know.
(but I am 1-in-a-million atypical on this)
So like, the formula for concentrations goes like this:
If we add 1volume of xconcentration to 2volume of yconcentration then:
we are adding 1volume of stuff to 2volume of stuff and getting 1volume+2volume of stuff.
We are adding xconcentration x 1volume to yconcentration x 2volume of special stuff and getting xconcentration x 1volume+yconcentration x 2volume of stuff
so our final concentration is the amount of special stuff divided by the amount of stuff, or
(xconcetration x 1volume+ yconcentration x 2volume)/(1volume+2volume)=zconcentration
So, I mean, that's all there is. That's the formula, that's how the problem works, that's all that's at issue.
It's equivalent to the problem:
You spend 2volume time at yconcentration speed. How much time must you spend at xconcentration speed to attain an average speed of zconcentration?