I had the opportunity a while back to wheel someone in and out of a home via a ramp. The ramp (one solid piece of aluminum) was much steeper (~15º) than the ADA guidelines (~5º). What I’m curious about is there a way of determining how much less effort it would have been to push under the standard guidelines. The numbers:
ADA recommended: 2’ rise with 24’ of ramp
What I had: 2’ rise with an 8’ ramp
My weight, the wheelchair, and the patient: ~350 pounds combined
TL;DR: Ignoring rolling friction the total effort would be the same. It’d just be done over 3x as long a duration, so 1/3rd the momentary effort.
Longer form:
The term “effort” isn’t defined in physics. There’s work, power, and energy, but not effort. And although human bodies are most definitely governed by physics, there is also the reality that they’re biological systems with more going on than simple levers and pulleys and … Plus our subjective sense of easy or hard is nowhere near linear.
The key term is “inclined plane”. With 3 times the slope as in your example you would have to push three times as hard (ignoring friction). So, assuming really good wheel bearings a 350 lbs load would take a 87.5 lb push on the steep ramp or 29.2 lbs on the recommended ramp.
It’s just the weight multiplied by the sine of the angle.
Note that, strictly speaking, a 15º ramp is not three times the slope of a 5º ramp. But since those are both small angles, it’s a pretty good approximation.
That’s probably about what I trying to ask. About how much is combined weight of the patient and the wheelchair pushing back against me. At 0, it’s not pushing back at all. At 90, if I were pushing them in a Bat-climb, for example, I’d have to be pushing 100% of the weight. So it looks. like in this case about 50 pounds.
It is what Napier said, the weight times the sine of the angle. Which for small angles is approximately the angle itself in radians. Of course, this ignores friction. It takes some effort to push a chair on level ground.