How to calculate the impact force of hitting a pothole?

I suspect its too complex to reasonably calculate - the tyre, bushings, subframe and suspension both absorb a bunch of the force (how much depends on a vast number of things like pressure, carcass design, damping, etc, all of which change with deperature) as does the road surface, whether the pothole is full of water/gravel, angle of the steering wheel, vertical movement of the car prior to the impact, etc etc etc.

you'd also have to figure out the decelleration rate of things in the car that aren't fixed, like fuel, drivetrain, passengers, luggage.

Short answer is lots.

Somewhere between and metric buttload and one and half metric buttloads. Depending on whether the impact is sufficient for the driver to spill their gin and tonic.

Surely a ballpark figure could be made, based on weight, speed, size of hole, basic/average suspension?

Or put a sensor on the far edge of the pothole and drive over it.
I'd be curious to see the results!

I tend to think of potholes in terms of the bike needed to ride over them.

- Wouldn't trouble a road bike
- Rigid mtb
- Could do with a Lauf
- XC bike
- Trail bike
- DH bike
- OMG NEED MOAR SUSPENSION

I think your parameters are correct, although I’d add tyre pressure, wheel diameter and tyre dimensions But it would be a complicated model to combine all those elements. Essentially what you are trying to calculate is the vertical movement of the wheel and the time this takes and combining these to get an acceleration and hence force. The complication is the tyre deflection has a big impact on the time The wheel has to deflect.

My guess is

60mp is 27m/s

The wheel deflects 10cm vertically on hitting a square edge that high.

Distance traveled whilst wheel is deflecting is about a wheel radius. Let’s guess 0.25m

Time to deflect is 0.25/27=0.009259 seconds

Let’s call it 0.01 seconds

Then to find the acceleration

S=0.5at^2

a=2s/t^2

a=2x0.1/0.01^2

a=2000 ms^2

For for force we multiply by mass

I’m going to guess the wheel disc and hub have a mass of 20kg

F=ma

F=2000x20=40,000N

My model doesn’t allow for deflection of the the car body or the force from the suspension

Wheel force transducers exist to measure all three axes of xyz force and mx my mz moments plus wheel speed at each wheel. They are rather expensive to buy / hire and are have some fairly clever real time calculations to separate the forces, somethinge the loadcells, compensate for rotational position etc.

Suspension arms, steering track rods and dampers etc can be fairly easily strain gauged and calibrated to measure axial load. I could tell you the type of loads, failure modes and how we investigate those on test rigs
but won't as that is customer / company IP 🙂.

@mick_r I thought my brilliant mathematical was going to get me a job offer 😉

I think the force of impact itself it quite straightforward. could be modelled a mass (the vehicle) with velocity, hitting a fixed edge. The other stuff, tyres, suspension, determines how that impact energy is dissipated through the vehicle.

Wow, thanks guys. The speed of deflection is staggering. Interesting numbers though.

"I think the force of impact itself it quite straightforward. could be modelled a mass (the vehicle) with velocity, hitting a fixed edge. "

What angle edge though.... vertical or some variation of slope would make a significant difference [to the speed of deflection] I would have thought.

I think your parameters are correct, although I’d add tyre pressure, wheel diameter and tyre dimensions But it would be a complicated model to combine all those elements. Essentially what you are trying to calculate is the vertical movement of the wheel and the time this takes and combining these to get an acceleration and hence force. The complication is the tyre deflection has a big impact on the time The wheel has to deflect.

My guess is

60mp is 27m/s

The wheel deflects 10cm vertically on hitting a square edge that high.

Distance traveled whilst wheel is deflecting is about a wheel radius. Let’s guess 0.25m

Time to deflect is 0.25/27=0.009259 seconds

Let’s call it 0.01 seconds

Then to find the acceleration

S=0.5at^2

a=2s/t^2

a=2×0.1/0.01^2

a=2000 ms^2

For for force we multiply by mass

I’m going to guess the wheel disc and hub have a mass of 20kg

F=ma

F=2000×20=40,000N

My model doesn’t allow for deflection of the the car body or the force from the suspension

looks impressive. no idea if it is correct, but would the fact the wheel is spinning impact things at all - gyroscopic effects turning/leaning it for sure, dunno if it changes a rotational impact

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I think the force of impact itself it quite straightforward. could be modelled a mass (the vehicle) with velocity, hitting a fixed edge. The other stuff, tyres, suspension, determines how that impact energy is dissipated through the vehicle.

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No the force in an impact is effected by how things move in the impact. Basically the longer the time you have for the object to change it’s motion the lower the force

I’m making it up as I go along on suspension but I’m confident on crumple zones. Imagine driving into a pillar of a motorway bridge. The car will stop. If you’re in car with a ridged chase then maybe it’ll stop in 5 cm. If you’re in a car with a crumple zone that stops in 30cm then the collision lasts 6 times as long and the force on the car is 6 times smaller

When rear wheel of a bike hits a square edge step up like a curb we know the force on the bike is less with rear suspension. Basically the rear suspension gives the bike more time to gain the upward velocity to move up that small distance. More time force

I don’t think the rotation makes much difference. The track of the wheel and other suspension characteristics will

But let’s remember this is me, A-level physics teacher vs Mick running a major test facility working in this area

You'll be better at the maths than me John 🙂. And I'm only small fry compared to certain guys at vehicle manufacturers.

It honestly depends on so much. Are they braking? Is the pothole long enough so the wheel has time to fall fully in then more like a kerb strike as it exits? Does the spring-damper fully bottom out the spring aid which creates really big and fast peak loads?

Being a physicist I went for the simplest option. Which will the cause an over estimated in this case. There is no step down just a step up. But as you say once you’ve bottomed out the suspension then your lifting the whole car and the forces are much higher and we know this can happen

worth a watch for seeing how everything moves during a pothole impact - 4.30 in to skip the talking. Its amazing how much non-vertical deflection of the wheels there is

Quoted When rear wheel of a bike hits a square edge step up like a curb we know the force on the bike is less with rear
suspension. Basically the rear suspension gives the bike more time to gain the upward velocity to move up that small
distance. More time force

You are correct but that is not a change in force of impact just the enrgy of impact is dissapated via a different mechanism, i.e the spring absorbs some of the energy and the wheel can deflect.

Think about hitting a very soft ball vs a very hard ball with a bat. the force of contact between the bat and ball does not really change but the transfer of energy or Impulse does, the soft ball will be in contact for a longer time and deform to absorb energy. The hard ball will be in contact for a short time and experience a greater change in momentum.

quoted What angle edge though…. vertical or some variation of slope would make a significant difference [to the speed of deflection] I would have thought.

It is and edge so the angle is not that important. the deflection and angle of deflection all happen after the impact. at worst we would be calculating a maximum force of impact. what alot of people here are talking about is energy dissapation or Impulse (time of contact) not the actual Force of impact which is what the OP asked.

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Think about hitting a very soft ball vs a very hard ball with a bat. the force of contact between the bat and ball does not really change but the transfer of energy or Impulse does, the soft ball will be in contact for a longer time and deform to absorb energy. The hard ball will be in contact for a short time and experience a greater change in momentum.

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In your example do the hard and soft ball move off at the same speed? The impulse is equal to the change of momentum and again equal to force x time. So if the balls moves off at the same speed then the soft ball has a lower force for greater time.

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You are correct but that is not a change in force of impact just the enrgy of impact is dissapated via a different mechanism, i.e the spring absorbs some of the energy and the wheel can deflect.

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When you watch the slow mo videos of bikes with flat landing don’t the hard tails squash the rear tyre more?

Treat it as 3 systems, the tyre, the unsprung mass and the sprung mass.

Treat the deformation of the tyre as having zero mass and frictionless.

Assuming you flatten the tyre but don't hit the rim. That means the contact patch is momentarily lets say equal to the rim diameter (ish, because it depends on the aspect ratio and rims size) x the wheel width.

So for an average family car ~16x8, times tyre pressure 35psi = 4,500lb, 2,250kg, 22,500N.

You'd then need to know spring rates, damping curves etc to figure out how much of that gets transmitted to the car. And you can argue about the speed of the car, size of the pothole, etc, but it's all just assumptions at the end of the day so somewhere between 20kN and the 40kN by Ossify is probably about right as a "any more and you're not driving home" figure.

I like that argument. I was thinking that one could do something similar with the stiffness of a suspension spring. But in a bad hit we know that force needed bottom out the suspension is exceeded

Assuming you flatten the tyre but don’t hit the rim. That means the contact patch is momentarily lets say equal to the rim diameter

I don't think that's correct. the pothole is only going to be pressing in a small amount of contact, not the entire surface area of the rim.

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In your example do the hard and soft ball move off at the same speed? The impulse is equal to the change of momentum and again equal to force x time. So if the balls moves off at the same speed then the soft ball has a lower force for greater time.

They are hit with the same force. why would they move off at the same speed? as the transfer of energy is lower for the soft ball

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When you watch the slow mo videos of bikes with flat landing don’t the hard tails squash the rear tyre more?

The mass of the rider and bike are the same (very similar) so hit the ground with the same force. However how the energy is absorbed by the two different bikes and how much the rider feels is different.

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The mass of the rider and bike are the same (very similar) so hit the ground with the same force. However how the energy is adsorbed by the two different bikes and how much the rider feels is different.

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No the force isn’t the same, the impulse is the same. The impulse is the change in momentum of the bike and rider. The impulse is also force x time. So the force on the FS bike lower as the time to slow down is greater.

An emergency stop is more force than a less aggressive stop?

I can construct an energy argument for the force being Lower on the FS bike if it helps. But generally the energy arguments are harder.

To be honest all of your basic calcs haven't been unreasonable in the 20-40kN range depending on how severe the impact. We'll typically use 50kN actuators for up to 3.5 tonne vehicle "4 poster" tests, and wfts typically come in those ranges for light-medium-heavy passenger cars:
https://www.kistler.com/GB/en/cp/wheel-force-transducers-for-cars-roadyn-s6-for-passenger-cars/P0001251

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That ballpark calc up there ^^ looks a good starting estimate and tells me two things:

1. I'm glad my car has tyres to try to dampen out some of the impact from potholes.

2. Hitting a kerb a 60 mph is going to hurt! I'm glad most potholes aren't quite that bad.

I recently measured the impact force of hitting a pothole with my face. I can confirm it seems like quite a bit of force. Luckily, my employment as a "before" model for whitening toothpaste adverts hasn't been affected.

So do those wheel force transducers actually replace the wheel and have the tyre mounted on them or bolt to the wheel tyre assembly? Never knew such things existed.

In that video above, the rear tyre had a hole punched in it, clearly visible as a lump about the size of a golf ball. I had two alloy wheels trashed on the inside rims by hitting the edge of a kerbside pothole, the impact levering the rim away from the tyre bead.
And let’s not forget that any number of cyclists have either been killed or injured by hitting potholes filled with water on wet roads, the impact collapsing the wheel rims.
I saw a photo yesterday of three racing cyclists on the cobbled section of a road race, their rims were literally folded with a crease in them.

There has also to be an optimally bad speed to hit a pothole.

At very slow speeds (cm/s slow) the wheel will roll in and climb out again without any impact. It'll hit the upslope full on but with so little kinetic energy it won't really register.

At very high speed (and obvs depends on width of pothole) gravity and the spring extension will barely have moved the wheel down and so very little impact on the wheel as it hits the other side even though you're carrying a lot of KE in - you've effectively jumped it.

Somewhere between you're carrying a lot of energy in, at a speed where the wheel can reach the bottom of the hole for maximum impact as you reach the other side.