4 ms·
> The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic ene
by Swizec 3mo ago
> The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71
But if the cars produce downforce this is no longer true because you brake harder (more friction available) at higher speeds!
This is how F1 cars pull 4G when breaking. Some custom cars (like one of Ken Block’s last monsters or the Valkyre) use active aero braking to even greater effect.
- tracerbulletx 3mo agoBut what if the cars are spherical cows?
- BigTTYGothGF 3mo agoCows can't roll that fast.
- zdragnar 3mo agoNot with that attitude
- deleted 3mo ago[deleted]
- elromulous 3mo agoOr on shabbos
- lstodd 3mo agoIt takes a stupid cow, but when they can climb mountains as they do it is not inconcievable that one can roll down. (I was suprised to see a cow jumping up on a ~3m rock ledge like it was nothing)
- ndsipa_pomu 3mo agoHowever, they can go up stairs, but not down them.
- lstodd 3mo agoConsidering that where I saw it they were doing that unsupervised since at least ~200BC (this being the Illyricum and then Dalmatia), I don't think they have any problems climbing down now.
- ordu 3mo agoSpherical cows on the other hand have to move with orbital velocity at least, or they fail to stay in vacuum for long and splash.
- dotancohen 3mo agoCows don't roll - they are frictionless so just glide over the surface smoothly.
- kibwen 3mo agoI'm sorry to inform you that those cows are going to have a hard time braking on that frictionless surface.
- PaulDavisThe1st 3mo agoBased on a hike in the Carson National Forest 2 days ago, the only reason a cow is on a frictionless surface is that the cow shat all over it.
- chrisweekly 3mo ago1. +1 insightful, thanks for sharing your physics knowledge 2. I know you know this, but for the sake of others, it's when _braking_ (applying the brakes), not _breaking_ (becoming broken). I'm not a pedant. But these errors jump out at me and I'm always a bit surprised and dismayed at this dichotomy; in our field, somehow the requisite attention to detail, the precision inherent to communicating scientific concepts, code, algorithms and formulae, is so often just abandoned when it comes to prose.
- spockz 3mo agoWhile it is true that some cars can brake harder due to downforce etc, the point from GP was that both cars brake/ decelerate at the same rate. Regardless of how exactly that deceleration is achieved.
- Swizec 3mo ago> the point from GP was that both cars brake/ decelerate at the same rate Point is that’s not always true. If they are the same type of car, and the car happens to be the kind with downforce, then their rate of deceleration greatly depends on air speed. A downforce car decelerates faster at higher speeds. This is why you often see race cars lock their wheels towards the end of the braking zone, never at the beginning. The driver has to release the brakes as the car decelerates because there’s less friction available. You go from pulling 4G at the beginning of the braking zone to pulling the usual 1G once your speed drops enough for downforce to become negligible. Alos! Many non-race cars actualy produce lift. Meaning the faster car decelerates at a slower rate than the slower car (0.8G vs 1G), making the effect from OP even more pronounced.
- sokoloff 3mo ago> This is why you often see race cars lock their wheels towards the end of the braking zone, never at the beginning. That’s not the only reason, and I’m not even sure it’s the majority reason. Braking in a straight line offers more braking traction than braking while turning. What happens towards the end of a braking zone? The turn in. (Which also shifts weight to the outside tire and away from the inside tire.)