Why does my bumper scrape when driving fast over a bump but not slow?
up vote
26
down vote
favorite
When driving my car over a bump or a quick change in gradient (from flat to uphill or downhill to flat), if I don't drive slowly and the bump/change in gradient is large enough, I'll end up with my front bumper dragging on the road/bump. But if I go slowly, there's no such issue.
I would expect that the wheels are still taking the same path regardless of the speed, so the car's body should also act the same regardless. Why does it make a difference when speed is changed? Why doesn't the car behave the same way regardless?
newtonian-mechanics everyday-life collision spring
New contributor
add a comment |
up vote
26
down vote
favorite
When driving my car over a bump or a quick change in gradient (from flat to uphill or downhill to flat), if I don't drive slowly and the bump/change in gradient is large enough, I'll end up with my front bumper dragging on the road/bump. But if I go slowly, there's no such issue.
I would expect that the wheels are still taking the same path regardless of the speed, so the car's body should also act the same regardless. Why does it make a difference when speed is changed? Why doesn't the car behave the same way regardless?
newtonian-mechanics everyday-life collision spring
New contributor
youtube.com/watch?v=UB9gA9Af5Tw
– Jasper
2 days ago
add a comment |
up vote
26
down vote
favorite
up vote
26
down vote
favorite
When driving my car over a bump or a quick change in gradient (from flat to uphill or downhill to flat), if I don't drive slowly and the bump/change in gradient is large enough, I'll end up with my front bumper dragging on the road/bump. But if I go slowly, there's no such issue.
I would expect that the wheels are still taking the same path regardless of the speed, so the car's body should also act the same regardless. Why does it make a difference when speed is changed? Why doesn't the car behave the same way regardless?
newtonian-mechanics everyday-life collision spring
New contributor
When driving my car over a bump or a quick change in gradient (from flat to uphill or downhill to flat), if I don't drive slowly and the bump/change in gradient is large enough, I'll end up with my front bumper dragging on the road/bump. But if I go slowly, there's no such issue.
I would expect that the wheels are still taking the same path regardless of the speed, so the car's body should also act the same regardless. Why does it make a difference when speed is changed? Why doesn't the car behave the same way regardless?
newtonian-mechanics everyday-life collision spring
newtonian-mechanics everyday-life collision spring
New contributor
New contributor
edited yesterday
Qmechanic♦
99.3k121781105
99.3k121781105
New contributor
asked 2 days ago
scohe001
242211
242211
New contributor
New contributor
youtube.com/watch?v=UB9gA9Af5Tw
– Jasper
2 days ago
add a comment |
youtube.com/watch?v=UB9gA9Af5Tw
– Jasper
2 days ago
youtube.com/watch?v=UB9gA9Af5Tw
– Jasper
2 days ago
youtube.com/watch?v=UB9gA9Af5Tw
– Jasper
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
up vote
26
down vote
accepted
I think that physics.SE deserves an answer in physics terminology.
Change of direction = acceleration. Acceleration is caused by force. Force is applied through suspension and compresses it.
Larger speeds lead to larger acceleration and larger force which in turn compresses the spring more.
To be a bit more precise, we could model the upwards curve as consisting of arc segments. It would mean that at the segment with curvature radius $r$ the acceleration is $v^2/r$ and the force is $mv^2/r$. If we suppose the spring is linear, the compression will equal force over the spring constant - $frac{mv^2}{rk}$.
Disclaimer: the spring is probably not linear, it should become increasingly high-degree-polynomial near the extremes of it's range (i.e. it should try to stop itself from compressing too much). The $m$ is not the mass of the car but something like the portion that rests on the axis in question. Which depends on multiple factors including the road curvature and the absolute acceleration of car. And the $r$ is not the curvature radius of the road but that of the car's body. Suspension is intended exactly to make it less than the curvature of the road - smoothen the bump.
3
Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false
– user2813274
yesterday
Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph.
– scohe001
yesterday
One thing that isn't clear in this answer is why there's a force pushing the car down. It's not simply because of the speed: it's because of air resistance, and the fact that the car is aerodynamically designed to convert that friction into a downward force.
– jpaugh
13 hours ago
1
@jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation.
– whatsisname
12 hours ago
1
This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump.
– MichaelK
2 hours ago
|
show 2 more comments
up vote
64
down vote
Because your car has a suspension.
A car's wheels are not rigidly attached to the frame. Rather, they are attached with springs and shock absorbers that allow the wheels to move somewhat relative to the rest of the car. This is generally speaking a good thing, as it means that when the wheels go over a small bump or pothole in the road, the frame does not necessarily need to move up and down; and so you, the passenger, get a smoother ride. But if the bumps are large and the speeds are too high, this same "smoothing" effect means that the frame of the car can come into contact with the road surface.
As to why this smoothing effect occurs: imagine that you have, on a table, a large block (standing in for the car), attached by a spring (the suspension) to a smaller block (the wheel.) Suppose that the table is very smooth, so that we can ignore friction between the table and the smaller blocks. If you pull the small block quickly & suddenly away from the larger block, the spring stretches a lot and the large block won't move very much at all: it has a lot of inertia, and so it can't accelerate very quickly. This is the equivalent of going over a bump or pothole at high speed: the wheel suddenly moves up or down relative to the frame, but the frame doesn't move up or down much at all.
But if you pull the small block away from the large block slowly, then the large block will follow the small block, while the spring doesn't stretch terribly much. In this case, the low acceleration of the large mass takes place over a longer time, and so it can move more while the force is being exerted on it. This is the equivalent of going over a bump/pothole at low speed; since the wheels move up or down relatively slowly, the frame of the car will follow them. If you go over a bump at low speed, this means that the frame will follow the wheels (which follow the road surface), rather than moving in something resembling a straight line and possibly hitting the road surface.
2
Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks.
– Nuclear Wang
yesterday
I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!!
– scohe001
yesterday
No, it's because the car has downforce. The suspension allows movement, but doesn't cause it.
– jpaugh
13 hours ago
2
@jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of.
– whatsisname
12 hours ago
Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers.
– Andrew Morton
6 hours ago
|
show 2 more comments
up vote
7
down vote
In order to avoid the issue, your car needs to be lifted over the bump. In the case of fast motion, this has to occur in a shorter time span, hence with a higher acceleration, hence with greater force - and this force has to be transferred from the ground via your wheels to the car body. However, the positional relation between your tires and the car body is governed by springs and shock absorbers, which are compressed under strong forces. Additionally, your tire bottom may be compressed for the same reason. It follows that the car body moves closer to the bump than it would in slow motion.
You could avoid this phenomenon by getting rid of all springs and dampers (and also the air-filled tires) and replacing them with rigid connections. In that case, the car body would be accelerated quite suddenly with great force at the bump - probably resulting in structural failure! There's a reason why these elastic elements are built into your car after all ...
add a comment |
up vote
1
down vote
Shocks have time factor. They will absorb more if the change is rapid. This is by design as your body and car do not want a rapid impact.
I believe springs also have a time factor.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
26
down vote
accepted
I think that physics.SE deserves an answer in physics terminology.
Change of direction = acceleration. Acceleration is caused by force. Force is applied through suspension and compresses it.
Larger speeds lead to larger acceleration and larger force which in turn compresses the spring more.
To be a bit more precise, we could model the upwards curve as consisting of arc segments. It would mean that at the segment with curvature radius $r$ the acceleration is $v^2/r$ and the force is $mv^2/r$. If we suppose the spring is linear, the compression will equal force over the spring constant - $frac{mv^2}{rk}$.
Disclaimer: the spring is probably not linear, it should become increasingly high-degree-polynomial near the extremes of it's range (i.e. it should try to stop itself from compressing too much). The $m$ is not the mass of the car but something like the portion that rests on the axis in question. Which depends on multiple factors including the road curvature and the absolute acceleration of car. And the $r$ is not the curvature radius of the road but that of the car's body. Suspension is intended exactly to make it less than the curvature of the road - smoothen the bump.
3
Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false
– user2813274
yesterday
Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph.
– scohe001
yesterday
One thing that isn't clear in this answer is why there's a force pushing the car down. It's not simply because of the speed: it's because of air resistance, and the fact that the car is aerodynamically designed to convert that friction into a downward force.
– jpaugh
13 hours ago
1
@jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation.
– whatsisname
12 hours ago
1
This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump.
– MichaelK
2 hours ago
|
show 2 more comments
up vote
26
down vote
accepted
I think that physics.SE deserves an answer in physics terminology.
Change of direction = acceleration. Acceleration is caused by force. Force is applied through suspension and compresses it.
Larger speeds lead to larger acceleration and larger force which in turn compresses the spring more.
To be a bit more precise, we could model the upwards curve as consisting of arc segments. It would mean that at the segment with curvature radius $r$ the acceleration is $v^2/r$ and the force is $mv^2/r$. If we suppose the spring is linear, the compression will equal force over the spring constant - $frac{mv^2}{rk}$.
Disclaimer: the spring is probably not linear, it should become increasingly high-degree-polynomial near the extremes of it's range (i.e. it should try to stop itself from compressing too much). The $m$ is not the mass of the car but something like the portion that rests on the axis in question. Which depends on multiple factors including the road curvature and the absolute acceleration of car. And the $r$ is not the curvature radius of the road but that of the car's body. Suspension is intended exactly to make it less than the curvature of the road - smoothen the bump.
3
Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false
– user2813274
yesterday
Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph.
– scohe001
yesterday
One thing that isn't clear in this answer is why there's a force pushing the car down. It's not simply because of the speed: it's because of air resistance, and the fact that the car is aerodynamically designed to convert that friction into a downward force.
– jpaugh
13 hours ago
1
@jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation.
– whatsisname
12 hours ago
1
This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump.
– MichaelK
2 hours ago
|
show 2 more comments
up vote
26
down vote
accepted
up vote
26
down vote
accepted
I think that physics.SE deserves an answer in physics terminology.
Change of direction = acceleration. Acceleration is caused by force. Force is applied through suspension and compresses it.
Larger speeds lead to larger acceleration and larger force which in turn compresses the spring more.
To be a bit more precise, we could model the upwards curve as consisting of arc segments. It would mean that at the segment with curvature radius $r$ the acceleration is $v^2/r$ and the force is $mv^2/r$. If we suppose the spring is linear, the compression will equal force over the spring constant - $frac{mv^2}{rk}$.
Disclaimer: the spring is probably not linear, it should become increasingly high-degree-polynomial near the extremes of it's range (i.e. it should try to stop itself from compressing too much). The $m$ is not the mass of the car but something like the portion that rests on the axis in question. Which depends on multiple factors including the road curvature and the absolute acceleration of car. And the $r$ is not the curvature radius of the road but that of the car's body. Suspension is intended exactly to make it less than the curvature of the road - smoothen the bump.
I think that physics.SE deserves an answer in physics terminology.
Change of direction = acceleration. Acceleration is caused by force. Force is applied through suspension and compresses it.
Larger speeds lead to larger acceleration and larger force which in turn compresses the spring more.
To be a bit more precise, we could model the upwards curve as consisting of arc segments. It would mean that at the segment with curvature radius $r$ the acceleration is $v^2/r$ and the force is $mv^2/r$. If we suppose the spring is linear, the compression will equal force over the spring constant - $frac{mv^2}{rk}$.
Disclaimer: the spring is probably not linear, it should become increasingly high-degree-polynomial near the extremes of it's range (i.e. it should try to stop itself from compressing too much). The $m$ is not the mass of the car but something like the portion that rests on the axis in question. Which depends on multiple factors including the road curvature and the absolute acceleration of car. And the $r$ is not the curvature radius of the road but that of the car's body. Suspension is intended exactly to make it less than the curvature of the road - smoothen the bump.
answered yesterday
Džuris
1,92711329
1,92711329
3
Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false
– user2813274
yesterday
Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph.
– scohe001
yesterday
One thing that isn't clear in this answer is why there's a force pushing the car down. It's not simply because of the speed: it's because of air resistance, and the fact that the car is aerodynamically designed to convert that friction into a downward force.
– jpaugh
13 hours ago
1
@jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation.
– whatsisname
12 hours ago
1
This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump.
– MichaelK
2 hours ago
|
show 2 more comments
3
Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false
– user2813274
yesterday
Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph.
– scohe001
yesterday
One thing that isn't clear in this answer is why there's a force pushing the car down. It's not simply because of the speed: it's because of air resistance, and the fact that the car is aerodynamically designed to convert that friction into a downward force.
– jpaugh
13 hours ago
1
@jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation.
– whatsisname
12 hours ago
1
This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump.
– MichaelK
2 hours ago
3
3
Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false
– user2813274
yesterday
Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false
– user2813274
yesterday
Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph.
– scohe001
yesterday
Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph.
– scohe001
yesterday
One thing that isn't clear in this answer is why there's a force pushing the car down. It's not simply because of the speed: it's because of air resistance, and the fact that the car is aerodynamically designed to convert that friction into a downward force.
– jpaugh
13 hours ago
One thing that isn't clear in this answer is why there's a force pushing the car down. It's not simply because of the speed: it's because of air resistance, and the fact that the car is aerodynamically designed to convert that friction into a downward force.
– jpaugh
13 hours ago
1
1
@jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation.
– whatsisname
12 hours ago
@jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation.
– whatsisname
12 hours ago
1
1
This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump.
– MichaelK
2 hours ago
This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump.
– MichaelK
2 hours ago
|
show 2 more comments
up vote
64
down vote
Because your car has a suspension.
A car's wheels are not rigidly attached to the frame. Rather, they are attached with springs and shock absorbers that allow the wheels to move somewhat relative to the rest of the car. This is generally speaking a good thing, as it means that when the wheels go over a small bump or pothole in the road, the frame does not necessarily need to move up and down; and so you, the passenger, get a smoother ride. But if the bumps are large and the speeds are too high, this same "smoothing" effect means that the frame of the car can come into contact with the road surface.
As to why this smoothing effect occurs: imagine that you have, on a table, a large block (standing in for the car), attached by a spring (the suspension) to a smaller block (the wheel.) Suppose that the table is very smooth, so that we can ignore friction between the table and the smaller blocks. If you pull the small block quickly & suddenly away from the larger block, the spring stretches a lot and the large block won't move very much at all: it has a lot of inertia, and so it can't accelerate very quickly. This is the equivalent of going over a bump or pothole at high speed: the wheel suddenly moves up or down relative to the frame, but the frame doesn't move up or down much at all.
But if you pull the small block away from the large block slowly, then the large block will follow the small block, while the spring doesn't stretch terribly much. In this case, the low acceleration of the large mass takes place over a longer time, and so it can move more while the force is being exerted on it. This is the equivalent of going over a bump/pothole at low speed; since the wheels move up or down relatively slowly, the frame of the car will follow them. If you go over a bump at low speed, this means that the frame will follow the wheels (which follow the road surface), rather than moving in something resembling a straight line and possibly hitting the road surface.
2
Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks.
– Nuclear Wang
yesterday
I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!!
– scohe001
yesterday
No, it's because the car has downforce. The suspension allows movement, but doesn't cause it.
– jpaugh
13 hours ago
2
@jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of.
– whatsisname
12 hours ago
Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers.
– Andrew Morton
6 hours ago
|
show 2 more comments
up vote
64
down vote
Because your car has a suspension.
A car's wheels are not rigidly attached to the frame. Rather, they are attached with springs and shock absorbers that allow the wheels to move somewhat relative to the rest of the car. This is generally speaking a good thing, as it means that when the wheels go over a small bump or pothole in the road, the frame does not necessarily need to move up and down; and so you, the passenger, get a smoother ride. But if the bumps are large and the speeds are too high, this same "smoothing" effect means that the frame of the car can come into contact with the road surface.
As to why this smoothing effect occurs: imagine that you have, on a table, a large block (standing in for the car), attached by a spring (the suspension) to a smaller block (the wheel.) Suppose that the table is very smooth, so that we can ignore friction between the table and the smaller blocks. If you pull the small block quickly & suddenly away from the larger block, the spring stretches a lot and the large block won't move very much at all: it has a lot of inertia, and so it can't accelerate very quickly. This is the equivalent of going over a bump or pothole at high speed: the wheel suddenly moves up or down relative to the frame, but the frame doesn't move up or down much at all.
But if you pull the small block away from the large block slowly, then the large block will follow the small block, while the spring doesn't stretch terribly much. In this case, the low acceleration of the large mass takes place over a longer time, and so it can move more while the force is being exerted on it. This is the equivalent of going over a bump/pothole at low speed; since the wheels move up or down relatively slowly, the frame of the car will follow them. If you go over a bump at low speed, this means that the frame will follow the wheels (which follow the road surface), rather than moving in something resembling a straight line and possibly hitting the road surface.
2
Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks.
– Nuclear Wang
yesterday
I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!!
– scohe001
yesterday
No, it's because the car has downforce. The suspension allows movement, but doesn't cause it.
– jpaugh
13 hours ago
2
@jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of.
– whatsisname
12 hours ago
Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers.
– Andrew Morton
6 hours ago
|
show 2 more comments
up vote
64
down vote
up vote
64
down vote
Because your car has a suspension.
A car's wheels are not rigidly attached to the frame. Rather, they are attached with springs and shock absorbers that allow the wheels to move somewhat relative to the rest of the car. This is generally speaking a good thing, as it means that when the wheels go over a small bump or pothole in the road, the frame does not necessarily need to move up and down; and so you, the passenger, get a smoother ride. But if the bumps are large and the speeds are too high, this same "smoothing" effect means that the frame of the car can come into contact with the road surface.
As to why this smoothing effect occurs: imagine that you have, on a table, a large block (standing in for the car), attached by a spring (the suspension) to a smaller block (the wheel.) Suppose that the table is very smooth, so that we can ignore friction between the table and the smaller blocks. If you pull the small block quickly & suddenly away from the larger block, the spring stretches a lot and the large block won't move very much at all: it has a lot of inertia, and so it can't accelerate very quickly. This is the equivalent of going over a bump or pothole at high speed: the wheel suddenly moves up or down relative to the frame, but the frame doesn't move up or down much at all.
But if you pull the small block away from the large block slowly, then the large block will follow the small block, while the spring doesn't stretch terribly much. In this case, the low acceleration of the large mass takes place over a longer time, and so it can move more while the force is being exerted on it. This is the equivalent of going over a bump/pothole at low speed; since the wheels move up or down relatively slowly, the frame of the car will follow them. If you go over a bump at low speed, this means that the frame will follow the wheels (which follow the road surface), rather than moving in something resembling a straight line and possibly hitting the road surface.
Because your car has a suspension.
A car's wheels are not rigidly attached to the frame. Rather, they are attached with springs and shock absorbers that allow the wheels to move somewhat relative to the rest of the car. This is generally speaking a good thing, as it means that when the wheels go over a small bump or pothole in the road, the frame does not necessarily need to move up and down; and so you, the passenger, get a smoother ride. But if the bumps are large and the speeds are too high, this same "smoothing" effect means that the frame of the car can come into contact with the road surface.
As to why this smoothing effect occurs: imagine that you have, on a table, a large block (standing in for the car), attached by a spring (the suspension) to a smaller block (the wheel.) Suppose that the table is very smooth, so that we can ignore friction between the table and the smaller blocks. If you pull the small block quickly & suddenly away from the larger block, the spring stretches a lot and the large block won't move very much at all: it has a lot of inertia, and so it can't accelerate very quickly. This is the equivalent of going over a bump or pothole at high speed: the wheel suddenly moves up or down relative to the frame, but the frame doesn't move up or down much at all.
But if you pull the small block away from the large block slowly, then the large block will follow the small block, while the spring doesn't stretch terribly much. In this case, the low acceleration of the large mass takes place over a longer time, and so it can move more while the force is being exerted on it. This is the equivalent of going over a bump/pothole at low speed; since the wheels move up or down relatively slowly, the frame of the car will follow them. If you go over a bump at low speed, this means that the frame will follow the wheels (which follow the road surface), rather than moving in something resembling a straight line and possibly hitting the road surface.
edited yesterday
Roger Lipscombe
1033
1033
answered 2 days ago
Michael Seifert
14.4k22752
14.4k22752
2
Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks.
– Nuclear Wang
yesterday
I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!!
– scohe001
yesterday
No, it's because the car has downforce. The suspension allows movement, but doesn't cause it.
– jpaugh
13 hours ago
2
@jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of.
– whatsisname
12 hours ago
Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers.
– Andrew Morton
6 hours ago
|
show 2 more comments
2
Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks.
– Nuclear Wang
yesterday
I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!!
– scohe001
yesterday
No, it's because the car has downforce. The suspension allows movement, but doesn't cause it.
– jpaugh
13 hours ago
2
@jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of.
– whatsisname
12 hours ago
Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers.
– Andrew Morton
6 hours ago
2
2
Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks.
– Nuclear Wang
yesterday
Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks.
– Nuclear Wang
yesterday
I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!!
– scohe001
yesterday
I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!!
– scohe001
yesterday
No, it's because the car has downforce. The suspension allows movement, but doesn't cause it.
– jpaugh
13 hours ago
No, it's because the car has downforce. The suspension allows movement, but doesn't cause it.
– jpaugh
13 hours ago
2
2
@jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of.
– whatsisname
12 hours ago
@jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of.
– whatsisname
12 hours ago
Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers.
– Andrew Morton
6 hours ago
Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers.
– Andrew Morton
6 hours ago
|
show 2 more comments
up vote
7
down vote
In order to avoid the issue, your car needs to be lifted over the bump. In the case of fast motion, this has to occur in a shorter time span, hence with a higher acceleration, hence with greater force - and this force has to be transferred from the ground via your wheels to the car body. However, the positional relation between your tires and the car body is governed by springs and shock absorbers, which are compressed under strong forces. Additionally, your tire bottom may be compressed for the same reason. It follows that the car body moves closer to the bump than it would in slow motion.
You could avoid this phenomenon by getting rid of all springs and dampers (and also the air-filled tires) and replacing them with rigid connections. In that case, the car body would be accelerated quite suddenly with great force at the bump - probably resulting in structural failure! There's a reason why these elastic elements are built into your car after all ...
add a comment |
up vote
7
down vote
In order to avoid the issue, your car needs to be lifted over the bump. In the case of fast motion, this has to occur in a shorter time span, hence with a higher acceleration, hence with greater force - and this force has to be transferred from the ground via your wheels to the car body. However, the positional relation between your tires and the car body is governed by springs and shock absorbers, which are compressed under strong forces. Additionally, your tire bottom may be compressed for the same reason. It follows that the car body moves closer to the bump than it would in slow motion.
You could avoid this phenomenon by getting rid of all springs and dampers (and also the air-filled tires) and replacing them with rigid connections. In that case, the car body would be accelerated quite suddenly with great force at the bump - probably resulting in structural failure! There's a reason why these elastic elements are built into your car after all ...
add a comment |
up vote
7
down vote
up vote
7
down vote
In order to avoid the issue, your car needs to be lifted over the bump. In the case of fast motion, this has to occur in a shorter time span, hence with a higher acceleration, hence with greater force - and this force has to be transferred from the ground via your wheels to the car body. However, the positional relation between your tires and the car body is governed by springs and shock absorbers, which are compressed under strong forces. Additionally, your tire bottom may be compressed for the same reason. It follows that the car body moves closer to the bump than it would in slow motion.
You could avoid this phenomenon by getting rid of all springs and dampers (and also the air-filled tires) and replacing them with rigid connections. In that case, the car body would be accelerated quite suddenly with great force at the bump - probably resulting in structural failure! There's a reason why these elastic elements are built into your car after all ...
In order to avoid the issue, your car needs to be lifted over the bump. In the case of fast motion, this has to occur in a shorter time span, hence with a higher acceleration, hence with greater force - and this force has to be transferred from the ground via your wheels to the car body. However, the positional relation between your tires and the car body is governed by springs and shock absorbers, which are compressed under strong forces. Additionally, your tire bottom may be compressed for the same reason. It follows that the car body moves closer to the bump than it would in slow motion.
You could avoid this phenomenon by getting rid of all springs and dampers (and also the air-filled tires) and replacing them with rigid connections. In that case, the car body would be accelerated quite suddenly with great force at the bump - probably resulting in structural failure! There's a reason why these elastic elements are built into your car after all ...
edited yesterday
SiHa
1032
1032
answered 2 days ago
Hagen von Eitzen
23115
23115
add a comment |
add a comment |
up vote
1
down vote
Shocks have time factor. They will absorb more if the change is rapid. This is by design as your body and car do not want a rapid impact.
I believe springs also have a time factor.
add a comment |
up vote
1
down vote
Shocks have time factor. They will absorb more if the change is rapid. This is by design as your body and car do not want a rapid impact.
I believe springs also have a time factor.
add a comment |
up vote
1
down vote
up vote
1
down vote
Shocks have time factor. They will absorb more if the change is rapid. This is by design as your body and car do not want a rapid impact.
I believe springs also have a time factor.
Shocks have time factor. They will absorb more if the change is rapid. This is by design as your body and car do not want a rapid impact.
I believe springs also have a time factor.
answered 2 days ago
paparazzo
1,165511
1,165511
add a comment |
add a comment |
scohe001 is a new contributor. Be nice, and check out our Code of Conduct.
scohe001 is a new contributor. Be nice, and check out our Code of Conduct.
scohe001 is a new contributor. Be nice, and check out our Code of Conduct.
scohe001 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f440510%2fwhy-does-my-bumper-scrape-when-driving-fast-over-a-bump-but-not-slow%23new-answer', 'question_page');
}
);
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
youtube.com/watch?v=UB9gA9Af5Tw
– Jasper
2 days ago