Minimum spring force required to keep a car level

In summary, the conversation discusses the attempt to solve a problem regarding the possibility of a car sitting at a certain angle. The solution involves using the spring force equation to calculate the minimum spring force needed to keep the car from tipping over, and considering factors such as the weight distribution, center of gravity, and torque from gravity and the springs. The conversation also includes an algebra
  • #1
at570

Homework Statement


This is an attempt to solve a problem I asked about here https://www.physicsforums.com/threads/can-a-car-ever-sit-like-this.929453/

Homework Equations


F = kx spring force

The Attempt at a Solution


using the largest angle down it can get with the front spring compressed all the way.
distance from the center of gravity to the front wheels / distance between the wheels = weight on the front wheels
convert from kg to n and divide by 2 wheels = minimum spring force to keep this from happening
but if the spring force is strong enough to push it up at the farthest angle down, will it be stronger than the weight the entire way back? As the spring gets longer the force will decrease and as the body rotates back the weight on the front wheels will also decrease.
 
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  • #2
at570 said:

Homework Statement


This is an attempt to solve a problem I asked about here https://www.physicsforums.com/threads/can-a-car-ever-sit-like-this.929453/

Homework Equations


F = kx spring force

The Attempt at a Solution


using the largest angle down it can get with the front spring compressed all the way.
distance from the center of gravity to the front wheels / distance between the wheels = weight on the front wheels
convert from kg to n and divide by 2 wheels = minimum spring force to keep this from happening
but if the spring force is strong enough to push it up at the farthest angle down, will it be stronger than the weight the entire way back? As the spring gets longer the force will decrease and as the body rotates back the weight on the front wheels will also decrease.
Start from the other end. Consider the car body level, then subjected to a small perturbation (rotation).
You will likely need variables for car mass, spring constants, distance between springs and height of mass above rotation centre.
 
  • #3
Im using
weight of sprung mass 160 kg
spring constant I am using 6000 n/m = 6 n/mm all 4 springs are the same
distance between springs 1m
height of mass above rotation center - I am a little fuzzy on this one. I think for a car with straight vertical suspension its on the ground under the center of gravity when its upright. Height = 1m

Im trying someone pushes down on the front so the center of gravity moves to 51% on the front wheels and let's go.
weight on front wheels = 160 kg * .51 = 81.6 kg
81.6 kg *10 = 816 n
816 n / 2 wheels = 408 n spring force
408 / 6000 = .068 m spring compression
Im not sure where to go from here
 
  • #4
at570 said:
height of mass above rotation center
A small perturbation from the equilibrium position would depress one set of springs by the same extent that the other set would expand. So the centre of rotation is halfway between the top ends of the springs.
If the perturbation is through an angle Δθ, what is the restoring torque from the springs? The centre of mass will now not be above the centre of rotation. What torque (tending to increase the perturbation) comes from gravity?
 
  • #5
Im getting 0 restoring force
Im doing
51% on front wheels
weight on front wheels = 160 kg * .51 = 81.6 kg
81.6 kg *10 = 816 n
816 n * .5 m = 408 nm
weight on back wheels = 160 kg * .49 = 78.4 kg
78.4 kg * 10 = 784 n
784 n * .5 m = 392 nm
408 nm - 392 nm = 16 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .01 m offset from roll center = 16 nm
16 nm - 16 nm = 0

52% on front wheels
weight on front wheels = 160 kg * .52 = 83.2 kg
83.2 kg *10 = 832 n
832 n * .5 m = 416 nm
weight on back wheels = 160 kg * .48 = 76.8 kg
76.8 kg * 10 = 768 n
768 n * .5 m = 384 nm
416 nm - 384 nm = 32 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .02 m offset from roll center = 32 nm
32 nm - 32 nm = 0
 
  • #6
at570 said:
Im getting 0 restoring force
Im doing
51% on front wheels
weight on front wheels = 160 kg * .51 = 81.6 kg
81.6 kg *10 = 816 n
816 n * .5 m = 408 nm
weight on back wheels = 160 kg * .49 = 78.4 kg
78.4 kg * 10 = 784 n
784 n * .5 m = 392 nm
408 nm - 392 nm = 16 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .01 m offset from roll center = 16 nm
16 nm - 16 nm = 0

52% on front wheels
weight on front wheels = 160 kg * .52 = 83.2 kg
83.2 kg *10 = 832 n
832 n * .5 m = 416 nm
weight on back wheels = 160 kg * .48 = 76.8 kg
76.8 kg * 10 = 768 n
768 n * .5 m = 384 nm
416 nm - 384 nm = 32 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .02 m offset from roll center = 32 nm
32 nm - 32 nm = 0
The trouble with plugging in numbers straight away is that you cannot see what is going on. Do it all algebraically.
 
  • #7
l = length between front and back spring
lg = moment arm of gravity
w = weight of car
d = distance center of gravity has moved closer to front wheels
xf = compression distance of front spring
xb = compression distance of back spring
lsf = moment arm of front spring
lsb = moment arm of back spring
k = spring constant
m = mass of car
tg = torque from gravity
tsf = torque from front spring
tsb = torque from back spring
tr = restoring torque
t = overall torque

fg = mg
tg = fg x l
tg = mg x l

w = mg
wsf = wl / 2 + d
wsb = wl / 2 - d

fsf = wsf
fsb = wsb

fsf = wl / 2 + d
fsb = wl / 2 - d

tsf = (l / 2) x fsf
tsb = (l / 2) x fsb

tr = tsf - tsb
t = tr - tgI was wondering since the suspension is attached solidly to the body the suspension would rotate too when the body rotates and the front wheel would move closer to the center of gravity so the center of gravity wouldn't have to move as far to get the same weight on the front wheels. I am not sure if its needed though. I would do it like this

fsfo = spring force on the front springs on one spring
fsbo = spring force on the back springs on one spring
xl = compression of springs at level
xsfd = compression difference from level to angled for a spring
a = angle from a line from the roll center to the top of the front spring to a line from the roll center to the distance down the spring its compressed from rolling
h = height of roll center off the ground
dw = distance the front wheels have moved back
da = actual distance the center of gravity moved relative to the roll center

spring force
f = kx
x = f / k

fsfo = fsf / 2
fsbo = fsb / 2

xl = .067 m

xsf = fsf / k
xsb = fsb / k

xsfd = xsf- xl

tan(a) = xsfd / ( l / 2)
a = tan-1 (xsfd / ( l / 2) )

dw = tan(a) / h

da = d - dw
 
  • #8
I tried a new approach and I think it works better. If the body rotates angle θ, first get the original compression of the springs at level by doing m is the sprung mass and do fg = mg and divide by 2 to get the force on the front springs and / 2 to get the force of one front spring. Then use the spring equation f = kx to get xo the original compression at level. Then if it rotates by θ make a triangle from the roll center to the front spring. The angle at the roll center is θ and the opposite is xθ the amount the front spring compresses from the rotation. Get xf the compression of the front spring by doing xf = xo + xθ and get the force for the front spring by doing f = kxf. Get the torque from the 2 front springs by doing τf = ( ff * 2 ) × l1 and l1 is the distance from the roll center to the spring. Then make another triangle for the back spring. Do xb = xo - xθ since the back spring rotates by the same amount but up. Use fb = kxb for the force of one back spring. Do τb =
( fb * 2 ) × l2 and l2 is the distance from the roll center to the back spring.
Then make another triangle from the roll center to the center of gravity. The angle at the roll center is θ and the hypotnuse is h, the height above the roll center of the center of gravity when the cars level. Get the opposite side and that's l3, the moment arm for the force of gravity. Do fg = mg and then τg = fg × l3 to get torque from gravity.
Then do τs = τf - τb to get the torque from the springs. Then τ = τs - τg to get the overall torque.
I tried it and if everythings good there, gravity is stronger than the springs under the right conditions
 
  • #9
Can someone confirm this is the way to do it?
 
  • #10
at570 said:
I tried a new approach and I think it works better. If the body rotates angle θ, first get the original compression of the springs at level by doing m is the sprung mass and do fg = mg and divide by 2 to get the force on the front springs and / 2 to get the force of one front spring. Then use the spring equation f = kx to get xo the original compression at level. Then if it rotates by θ make a triangle from the roll center to the front spring. The angle at the roll center is θ and the opposite is xθ the amount the front spring compresses from the rotation. Get xf the compression of the front spring by doing xf = xo + xθ and get the force for the front spring by doing f = kxf. Get the torque from the 2 front springs by doing τf = ( ff * 2 ) × l1 and l1 is the distance from the roll center to the spring. Then make another triangle for the back spring. Do xb = xo - xθ since the back spring rotates by the same amount but up. Use fb = kxb for the force of one back spring. Do τb =
( fb * 2 ) × l2 and l2 is the distance from the roll center to the back spring.
Then make another triangle from the roll center to the center of gravity. The angle at the roll center is θ and the hypotnuse is h, the height above the roll center of the center of gravity when the cars level. Get the opposite side and that's l3, the moment arm for the force of gravity. Do fg = mg and then τg = fg × l3 to get torque from gravity.
Then do τs = τf - τb to get the torque from the springs. Then τ = τs - τg to get the overall torque.
I tried it and if everythings good there, gravity is stronger than the springs under the right conditions
That's the process, yes.
You did not describe the conditions under which the gravitational torque beats the restoring torque, but I would guess it's when the height to the mass centre is large in proportion to the distance between the springs.
 
  • #11
Yep. I want to make it into something where you can find out if your cars going to do this based on the set up. But I want to work on an extension of it first
 
  • #12
An extension of the question: The springs are solidly attached to the body so when the body rotates the springs also rotate. When the springs rotate angle θ theyre not pointing straight up anymore so the force is less than what's needed to oppose gravity. What happens to the springs and body when this happens?
The springs forces upward have to equal gravity. Make a triangle at the connection of a spring and the body. The angle at the connection is θ. The hypotnuse is the spring force and adjacent is the part of the spring force that opposes gravity. I think what happens is either the body rotates again or it goes straight down until all the spring forces equal gravity. I am not really sure where to go from here
 
  • #13
at570 said:
The springs are solidly attached to the body so when the body rotates the springs also rotate.
But how are they attached to the axles? Are they rigidly attached there or pivoted?
 
  • #14
Bystander said:
Google "what is a toggle mechanism."
See also "trip balance/roberval principle/steelyard (balance)."
 
  • #15
These particular ones are rigidly attached which gave me some trouble figuring out where the roll center is since they don't really have instant centers, or its at infinity.
Also I see about lever arms and stuff and how it has to do with this. Everything has to be in balance both mass and distance for it not to rotate
 
  • #16
at570 said:
These particular ones are rigidly attached
To both the chassis and the axles? So the springs bend as well as expand and contract. Not sure how that affects the effective spring constant. But within the normal range of movement, I would think you can ignore this.
 
  • #17
I think its like a solid connection with a spring, so the solid part keeps it pretty much straight and also let's the spring move and compress
 
  • #18
at570 said:
I think its like a solid connection with a spring, so the solid part keeps it pretty much straight and also let's the spring move and compress
If the spring stays straight then the ends must be pivoted.
 
  • #19
Why?
 
  • #20
at570 said:
Why?
Because the chassis rotates while the axles stay put. How is that possible if the springs are attached to fixed points on each, do not bend or pivot, and the chassis structure does not stretch?
 
  • #21
The wheels can rotate wouldn't that take the strain off of the axles?
 
  • #22
at570 said:
The wheels can rotate
That has nothing to do with the geometric relationship between the axles, the springs and the chassis.
 
  • #23
Im not really understanding. The way the suspension looks there's a tube that sort of slides that keeps it straight, with the spring and there's no other connections to the wheel and theyre independent
 
  • #24
Oh also the way I drew it in the other post I guess the springs actually are at an angle when the body is at an angle
 
  • #25
at570 said:
Im not really understanding. The way the suspension looks there's a tube that sort of slides that keeps it straight, with the spring and there's no other connections to the wheel and theyre independent
We can abstract it as two fixed points, A and B, at the lower ends of thesprings, with a rigid block attached to the top ends of the springs at C and D.
If we rotate the block, compressing one spring more than the other, the horizontal distance from C to D reduces but the horizontal distance from A to B remains constant. It follows that the springs AC and BD cannot both remain vertical. If a spring does not remain vertical then it is either pivoted at the lower end or bends.
 
  • #26
Oh. So if the brakes are on the friction of the tires would try to keep it from rotating and it couldn't rotate like that because it would get out of shape. But if the brakes are not on wouldn't the wheels at A and B be able to roll to keep it from getting out of shape?
 
  • #27
at570 said:
Oh. So if the brakes are on the friction of the tires would try to keep it from rotating and it couldn't rotate like that because it would get out of shape. But if the brakes are not on wouldn't the wheels at A and B be able to roll to keep it from getting out of shape?
I don't see what the rolling of the wheels has to do with it. They do not alter the geometrical relationship between the four points I defined.
 
  • #28
Also the car looks more like this
26hEiAH.png

But if the effect on the springs is small mabey we can ignore it?
 

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  • #29
Actually, wouldn't the wheels be like pivots at A and B? So we can treat A and B like pivots I think
 
  • #30
at570 said:
Actually, wouldn't the wheels be like pivots at A and B? So we can treat A and B like pivots I think
So the springs are pivoted. Whether the brakes are on is irrelevant.
Don't know why I thought the distance between the wheels was fixed - that was dumb.
 
Last edited:
  • #31
Oh I get it. By axle I was thinking the assembly that has the rotating piece that connects to the wheels inside of it. I think youre asking me about the rotating piece right? Or asking if the whole wheel/axle area acts like a pivot joint
So if the springs are pivoted at the bottom then the front one can move farther under relative to the body and the back one can move farther out. I think there would be force opposing gravity and I think it would be different for each one. You can find them by dividing the distance from one to the center of gravity by the distance of the other one to the center of gravity * mg. But I am not quite sure where to go from here. Mostly about how to find the spring compression
 
  • #32
Saw your edit. So I guess you do treat the wheels as moving freely then. I am still trying to figure out how to get the spring compression with the springs angled
 
  • #33
I don't think you can do it the same way anymore because of the wheels now being included and the springs angled. Make a triangle at the connection of one of the springs to the body. The angle the body is at is θ. The angle of the triangle at the spring connection is θ. The hypotnuse is the spring force and adjacent is the part of the spring force opposing gravity. All of the adjacents together have to equal the force of gravity. Should I do this by weight and would it be the amount of weight on each wheel since the wheel positions are also different because of the rotation, one has gone underneath and one has gone out, so that force transferred up the suspension to the spring, or by weight at the spring connection to the body? I would think it would be the force tranferred up the suspension. I wonder though if i did it by weight if youd always get zero rotational torque.
Or should I do it by angle, since the angle of rotation is θ. Id make a triangle halfway between the connections of the springs to the body to one connection of the spring to the body. The angle is θ, adjacent is half the distance between the connections of the springs to the body and the opposite is the spring compression from rotation. But the spring compression from rotation only isn't enough to equal gravity anymore because of the rotation of the springs. There needs to be some extra force and I am not sure if its distributed evenly or by percent of weight on each wheel or possibly another way.
Im also wondering because the wheels and body are both now angled, would the spring compressions and because of that the distance from the suspension to the bottom of the wheel for the two springs always be the same with that angle and mass? I would think so
Also has the roll center moved since the wheels and whole suspension is included now? Is it still halfway between the connections for the springs to the body or it it somewhere else mabey on the ground under the center of mass when level?
Can someone give me some ideas or comments on these ways of doing it?
 
  • #34
I saw that the rotation center of a car front to back is the pitch center. Make the first two triangles at level and then rotate it angle θ. Make a triangle from the pitch center to to connection between the body and the front spring. The height is so the height of the bottom of the body when level opposite is d1 the distance from the center of the body to the connection to the front spring. Get the angle at the pitch center a and the hypotnuse p1 the distance from the pitch center to the connection of the spring to the body. Make a triangle from the pitch center. The hypotnuse is p1 and the angle at the pitch center is a + θ. The adjacent is l4 the moment arm for the front spring force in the x direction and opposite is l1 the moment arm for the front spring force in the y. Make another triangle from the pitch center to the connection of the back spring to the body. The adjacent is so and the opposite is d1. Get the angle at the pitch center a and the hypotnuse p2 the distance from the pitch center to the connection of the back spring to the body. Make a triangle from the pitch center. The angle at the pitch center is a - θ and the adjacent is l2 the moment arm of the back spring force in the y and opposite is l5 the moment arm of the back spring force in the x. Make a triangle from where the front spring connects to the body down. Angle is θ and adjacent is l4. Get the hypotnuse sf the length of the front suspension to the bottom of the wheel now. Make a triangle from the connection of the back spring to the body down. The angle is θ and adjacent is l5. Get the hypotnuse sb the length of the back suspension to the bottom of the wheel now. Do so - sf = xfd difference in spring compression for front spring from level. Do xfd + xo = xf total front spring compression. xo is original spring compression at level. Do so - sb = xbd difference in spring compression for back spring from level. Do xbd + xo = xb total back spring compression. Do xf * k = ff front spring force. k is spring constant. Do xb * k = fb back spring force. Make a triangle. Angle is θ and hypotnuse is ff. Get adjacent is ffy front spring force in the y direction and opposite is ffx front spring force in the x. Make a triangle. Angle is θ and hypotnuse is fb. Get adjacent fby back spring force in the y and opposite fbx back spring force in the x. Do ( ffy * 2 ) × l1 = τfy front spring force in the y and ( fby * 2 ) × l2 = τby back spring torque in the y and ( ffx * 2 ) × l4 = τfx front spring torque in the x and ( fbx * 2 ) × l5 = τbx back spring torque in the x.
Make a triangle from the pitch center to the center of gravity. Angle is θ and hypotnuse is h the height of the center of gravity above the pitch center at level. Get the opposite l3 moment arm of gravity. Do fg = mg force of gravity. Do fg × l3 = τg torque from gravity.
Do τfy - τby - τfx - τbx = τs spring torque. Do τs - τg = τ overall torque
I tried it out a few times and if everythings good here with the right set up it has a place where gravity is stronger than the springs decreasing and then the springs are stronger than gravity increasing and it can get stuck there. I think the spring force increases because of the front wheel moving farther under the body. The set up is a small distance between the wheels and a high center of gravity compared to it.
Can someone confirm this is correct/the correct process?
 
  • #35
Thanks to everyone who helped out, esp. haruspex
 

What is the minimum spring force required to keep a car level?

The minimum spring force required to keep a car level depends on various factors such as the weight of the car, the stiffness of the springs, and the suspension design. Generally, a minimum force of 50-100 pounds per inch of compression is recommended for most cars.

How do you calculate the minimum spring force required for a specific car?

To calculate the minimum spring force required for a specific car, you will need to know the weight of the car, the spring rate (stiffness) of the springs, and the suspension geometry. You can use the formula F = kx, where F is the force, k is the spring rate, and x is the compression distance.

What happens if the minimum spring force is not met?

If the minimum spring force required to keep a car level is not met, the car's suspension will not be able to support the weight of the car, leading to sagging and poor handling. This can also cause damage to the suspension components and affect the overall performance of the car.

Can the minimum spring force be adjusted?

Yes, the minimum spring force can be adjusted by changing the stiffness of the springs or adjusting the suspension geometry. However, it is important to consult a professional and make adjustments carefully to ensure the car's stability and safety.

Is the minimum spring force the same for all cars?

No, the minimum spring force required to keep a car level may vary depending on the weight and design of the car. It is important to consider the specific factors of each car when determining the minimum spring force required.

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