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### #Actualbmarci

Posted 04 October 2012 - 04:37 AM

One more thought on suspensions (for dummies (mostly myself) ) through an example.

For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.

The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)

Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)

In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.

^^^ Could this work????

Couple of things to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just summing the spring and demper rates?

### #3bmarci

Posted 04 October 2012 - 04:36 AM

One more thought on suspensions (for dummies (mostly myself) ) through an example.

For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.

The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)

Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)

In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.

^^^ Could this work????

Couple of things to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just adding the spring and demper rates?

### #2bmarci

Posted 04 October 2012 - 04:33 AM

One more thought on suspensions (for dummies (mostly myself) ) through an example.

For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.

The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)

Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)

In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.

^^^ Could this work????

Couple of thing to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just adding the spring and demper rates?

### #1bmarci

Posted 04 October 2012 - 04:32 AM

One more thought on suspensions (for dummies (mostly myself) ) through an example.

For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.

The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)

Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)

In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
ratio=(m_wheel/m_car) or maybe (Wsprung/W0) ratio should be different for each wheel based on the original weight distribution
Fwheel=F*ratio and Fcar=F*(1-ratio)
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.

^^^ Could this work????

Couple of thing to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just adding the spring and demper rates?

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