this post was submitted on 16 Jul 2025
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[โ€“] HaraldvonBlauzahn@feddit.org 2 points 3 days ago* (last edited 3 days ago) (1 children)

It is very clear from statistics of traffic accidents between cars and pedestrians that risk of lethal injuries rises sharply with speed, even at speeds of 30 km/h. It does not make a difference whether a car crashes with 30 km\h into you, or you crash with 30 km/h into a car.

It is also very clear that riding light motorcycles is far more risky than riding a bike.

[โ€“] Alerian@sh.itjust.works 2 points 3 days ago (1 children)

Actually it does make a difference, as cynetic energy is proportionnal to mass, so 30kmh car is much more dangerous than a 30kmh bike. Though it does not invalidate your point

[โ€“] HaraldvonBlauzahn@feddit.org 2 points 3 days ago* (last edited 3 days ago) (1 children)

No, that's not the case. What's relevant, especially with a difference of mass so large, is the relative speed of the two objects, which is 30 km/h.

[โ€“] Successful_Try543@feddit.org 1 points 2 days ago (2 children)

@Alerian@sh.itjust.works

Suppose the impact coefficient k is similar, it does make a difference whether a bike crashes into a standing car (case 1) or a car crashes into a standing biker (case 2).

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[โ€“] HaraldvonBlauzahn@feddit.org 2 points 1 day ago (1 children)

The thing is: You are using velocities v1, v2 which are relative to Earth. But none of the two vehicles collide with Earth - they collide with each other, thus the thing that matters is their relative speed, thus the difference of their velocities relative to Earth.

(That's also why the speed at which both Earth, the car, and the motorized bike move around the sun does not matter - relative speed is all what matters).

The other thing is that a human colliding with an object of several tons weight with a speed of, say, 36 km/h is not "elastic". 36 km/h is 10 meter per second, which is equal to about one second of free fall (accelerating with a= 9.81 meter per square second to the ground), which is equivalent to a fall height of h = a/2 * s ^2 or 5 meters.

Somebody falling from 5 meters hight on hard concrete ground will not bounce up but will likely have some broken bones, or a broken skull. What happens is that all parts of thier body is decelerated to a speed of zero within a distance of one or two centimeters, which involves massive forces that easily break bones.

And a speed of 14 m/s, or 54 km/h corresponds to a fall of ten meters depth - almost certainly lethal if hitting a two-ton concrete block.

[โ€“] Successful_Try543@discuss.tchncs.de 1 points 23 hours ago* (last edited 23 hours ago)

The thing is: You are using velocities v1, v2 which are relative to Earth.

The formula includes the relative speed (v_2 - v_1) of both bodies. Derivation, see Wikipedia or a book on engineering mechanics.

Somebody falling from 5 meters hight on hard concrete ground will not bounce up but will likely have some broken bones, or a broken skull. What happens is that all parts of thier body is decelerated to a speed of zero within a distance of one or two centimeters, which involves massive forces that easily break bones.

case 1, k=0. Fortunately, a car is not solid rock. I don't know about a typical of k for collissions of humans with a car, but if you say it's 0, that's actually good for the biker, as the forces then acting on their tissues is smaller than if that would not the case.

So concluding. If the collision of the biker and the car is completely inelastic, it doesn't matter if the biker crashes into a resting car or the car crashes into a standing biker. The only thing that matters is the relative velocity of the two objects.

[โ€“] Alerian@sh.itjust.works 1 points 2 days ago (1 children)

Sorry I don't have much time today to get into it. Seems to me you can't solve for case 2 here since in this case m2 and m1 are switched. But it does not matter, I am not trying to solve for speed before and after.

The force of the impact does not depend on the mass, I agree, but the energy to dissipate (in the cyclist body) is much higher. I'm just saying that inertia plays a role as it contribute to the energy necessary to stop either vehicule. I am happy to be proven wrong, I just don't think this is the right equation to do so.

[โ€“] Successful_Try543@feddit.org 1 points 1 day ago* (last edited 1 day ago)

Seems to me you can't solve for case 2 here since in this case m2 and m1 are switched.

No, for both cases, body "1" is the biker and body "2" is the car.

The force of the impact does not depend on the mass, I agree,

This is just for the sake if simplicity. The force does in general depend on both masses, not just the mass of the car. Yet, the biker has only ~ 5 - 10 % of the mass of the car and thus, their mass can be neglected and the simplfied solution (m_1 / m_2 -> 0) doesn't include masses anymore

but the energy to dissipate (in the cyclist body) is much higher. I'm just saying that inertia plays a role as it contribute to the energy necessary to stop either vehicule.

Exactly. This part is included in the coefficient k. Yet, for the simplified solution, the biker doesn't stop the car in any form.

Suppose a completely plastic impact, k=0: The biker would be stopped to zero velocity in case 1, and in case 2 they would be accelerated to the velocity of the car. Here the magnitude of the force/acceleration doesn't depend on whether the bike did move or the car did move.

For the elastic case, k=1, car and bike are treated as billard balls: For case 1, the biker moves with the same velocity as before, but in opposite direction. For the other case, the biker would move in opposite direction, but with the double velocity as in case 1. Thus, here, the force causing the acceleration must also be twice.

So as long as the impact is not purely plastic, it does matter whether the biker hits the car (case 1) or the car hits the biker (case 2).