no ... it adds weight it doenst add mass and therfore doesnt add inertia ...
no ... it adds weight it doenst add mass and therfore doesnt add inertia ...
That may be true, but I'm pretty sure it does have an effect on suspension frequency in addition to travel.
Forgive me if this has already been discussed, but I just started looking into this spring frequency calculation stuff and stumbled upon something interesting. The formulas I've seen all discuss spring rate. However, shouldn't wheel rate be more important? I.e the wheel rate is the spring rate multiplied by the motion ratio. Since spring compression is usually at an angle relative to the motion of the wheel, the ideal spring rate will be higher than the ideal wheel rate.
Perhaps that is why RL rates seem to be lower than lfs? The RL values might be frequency related to wheel rate while the lfs calculators have been calcuating frequency based on spring rate alone? Here's a link to a page that gives formulas for calculating wheel rate based on suspension geometry: http://www.swayaway.com/Suspension%20Worksheet.htm
Any thoughts?
Perhaps that is why RL rates seem to be lower than lfs? The RL values might be frequency related to wheel rate while the lfs calculators have been calcuating frequency based on spring rate alone? Here's a link to a page that gives formulas for calculating wheel rate based on suspension geometry: http://www.swayaway.com/Suspension%20Worksheet.htm
Any thoughts?
Yeah, the effect of downforce on simple harmonic motion is a bit wierd. Havent quite got my head around all the effects yet.
It certainly doesnt add mass or inertia for obvious reasons, so the acceleration of the mass in response to given forces doesnt change. But it does add a whole load of force, in one direction only.
So the mass accelerates more quickly going downwards, and decellerates more quickly when moving upwards. Have to have a tinker with the old spreadsheets and see what I can come up with.
It certainly doesnt add mass or inertia for obvious reasons, so the acceleration of the mass in response to given forces doesnt change. But it does add a whole load of force, in one direction only.
So the mass accelerates more quickly going downwards, and decellerates more quickly when moving upwards. Have to have a tinker with the old spreadsheets and see what I can come up with.
The rates given in LFS are all wheel rates as far as we know, so this shouldnt be an issue.
Ok, so a 120 kN/m "spring" is really a 120 kN/m rate as it acts on the wheel, cool. I'm guessing then that the suspension geometries in lfs are linear with regards to motion ratio then. IRL, some suspensions have progressive or digressive ratios, which would make things complicated :eek:
Without going back into physics calculations, I'd say that downforce is equivalent to stronger gravitation (as long as the downforce doesn't change due to speed changes). I.e. constant force pointing down. IMHO, it will reduce the amplitude but not the frequency.
Here's an quote from here
Here's an quote from here
Exactamundo.
I plugged the extra downforce into the drop simulation graph in the old excel setup analyser, and took all the damping off to see undamped harmonic motion, and you basically get the exact same frequency with higher amplitude.
I plugged the extra downforce into the drop simulation graph in the old excel setup analyser, and took all the damping off to see undamped harmonic motion, and you basically get the exact same frequency with higher amplitude.
Really? Well that buggers up a lot of my guide then.
Ok, after little further investigation of this subject, I can declare that, under ideal (*) conditions:
1. constant downforce IS equivalent to greater gravity.
2. Any amount of gravity does NOT affect neither frequency NOR amplitude of the motion.
3. Therefore, the only thing that does change with the presence of downforce is the center of gravity, which is getting lower by additional F/K, where F is the downforce and K is the spring's stiffness factor.
(*) ideal conditions means that through the whole amplitude of the vertical motion, the spring stays at its linear-response range. This is not at all guarenteed under real conditions. If the introduction of additional downforce (or any other force caused by hitting a bump etc) causes the spring to get near the edge of it's range, then the motion gains "distortions" due to the non-linear response of the spring, and is no longer a simple harmonic motion. I think it's beyond me to calculate the motion in such cases
1. constant downforce IS equivalent to greater gravity.
2. Any amount of gravity does NOT affect neither frequency NOR amplitude of the motion.
3. Therefore, the only thing that does change with the presence of downforce is the center of gravity, which is getting lower by additional F/K, where F is the downforce and K is the spring's stiffness factor.
(*) ideal conditions means that through the whole amplitude of the vertical motion, the spring stays at its linear-response range. This is not at all guarenteed under real conditions. If the introduction of additional downforce (or any other force caused by hitting a bump etc) causes the spring to get near the edge of it's range, then the motion gains "distortions" due to the non-linear response of the spring, and is no longer a simple harmonic motion. I think it's beyond me to calculate the motion in such cases
Can you explain in detail how you come to this conclusion please?
Its clear to me that the overall gravity effects amplitude in a vertically oriented spring system. Downward force effects downward acceleration which results in higher velocity passing through the zero point, which results in longer overall extension to reverse the direction, which results in higher upward spring forces, which results in higher upward velocity passing the zero point, which results in higher maxima of the upward phase.
Ie. higher gravity results in higher amplitude.
I've added a G slider to the oscillation graph in the analyser and it clearly shows that increased G increases amplitude.
Now you can reasonably hold the opinion that an iterative graph generation is not the same thing as a mathematical proof, but nonetheless, its fairly clear.
Ok, I hope I'll be ok with the explanation.
1. The movement of the mass is only affected by the forces upon the mass at any given point (which affects the acceleration, etc).
2. Gravity affects the mass with constant force of F=m*g ( --> dependent on the mass)
3. downforce adds constant force (independent of the mass)
4. Gravity + constant force (per fixed mass) produce a single constant combined force that does not change during the movement of the mass.
5. if you combine the forces that "work" on the mass at any given point (i.e. adding the spring forces along with the gravity/downforce), you see that the result is exactly as if you've lowered the spring equilibrium point. The new equilibrium point is where K*x= g*m+F --> x= (g*m + F)/K where x is the displacement of the new equilibrium point relative to that of a no-external-forces point, g is gravitational constant, K is the spring's stiffness factor and F is the downforce.
I hope what I said makes any sense
1. The movement of the mass is only affected by the forces upon the mass at any given point (which affects the acceleration, etc).
2. Gravity affects the mass with constant force of F=m*g ( --> dependent on the mass)
3. downforce adds constant force (independent of the mass)
4. Gravity + constant force (per fixed mass) produce a single constant combined force that does not change during the movement of the mass.
5. if you combine the forces that "work" on the mass at any given point (i.e. adding the spring forces along with the gravity/downforce), you see that the result is exactly as if you've lowered the spring equilibrium point. The new equilibrium point is where K*x= g*m+F --> x= (g*m + F)/K where x is the displacement of the new equilibrium point relative to that of a no-external-forces point, g is gravitational constant, K is the spring's stiffness factor and F is the downforce.
I hope what I said makes any sense
That makes perfect sense, but all you've shown is that the spring equilibrium point is lower, which I agree with completely. What you've just worked through sheds no light on the issue of amplitude at all.
Okay, I'll try explaining it a different way. Please try and answer this riddle.
In system A, gravity = 9.8 m/s. When the mass is moving downwards on the spring, it is being accelerated by both the spring force and gravity, after it passes spring natural length is is being decellerated by spring force and accelerated by gravity, so its decelleration = g - spring decelleration, and maximum amplitude is reached when this force decellerates the mass to zero velocity.
In system B gravity = 19.6 m/s. As it passes zero point, it will have a higher velocity, as the downward force on it is greater. Yet you claim that it will decellerate to rest in the same distance as in System A, despite it having a higher velocity, and being decellerated at a lower rate due to the increased downward G force.
So the question is, where does the mystery force come from?
Okay, I'll try explaining it a different way. Please try and answer this riddle.
In system A, gravity = 9.8 m/s. When the mass is moving downwards on the spring, it is being accelerated by both the spring force and gravity, after it passes spring natural length is is being decellerated by spring force and accelerated by gravity, so its decelleration = g - spring decelleration, and maximum amplitude is reached when this force decellerates the mass to zero velocity.
In system B gravity = 19.6 m/s. As it passes zero point, it will have a higher velocity, as the downward force on it is greater. Yet you claim that it will decellerate to rest in the same distance as in System A, despite it having a higher velocity, and being decellerated at a lower rate due to the increased downward G force.
So the question is, where does the mystery force come from?
hehe, this turns out quite interesting. Maybe it's due to different initial assumptions and different perspective on what variables to consider and when.
So, I must admitt that your description seems logical, and I find it hard to invalidate it when following your steps. So perhaps you could solve the mystery following this new description along with the image:
the left spring represents the initial state of a mass on a spring without any forces. The right side represents the initial state when a constant force is applied.
Now, what I ment earlier, was that the total forces at any given point on the left spring around X0 are identical to these of the right spring around X1 (simple calculation of the sum of the forces the spring produce along with the constant external force on the axis of motion). Since the mass on both springs is identical, and the total forces around the relevant pivot point are identical as long as you're not on the edges of the spring's range, they MUST behave exactly the same around their relevant pivot point.
So, if you now apply the same force F0 for duration dT on both springs independantly (simulate a bump withoput taking into account the elevation of the car etc), they will aquire the same initial evergy and with only the total forces that affect them identical relative to their initial position, they will move exactly the same except for the x displacement.
So basicaly, a mass on a spring + constant force is identical to a mass on a spring without external forces, with the same K, but with different "Zero" point.
does this explains my oppinion better?
ps.
the numbers (-2 -1... 2) on both sides represent the total force on the mass along the axis of motion.
im pretty sure the amplitude depends on intitial deflection and not gravity or additional forces ...
Veeryy interesting.
It is very much a case of our assumptions and initial positions being different. Again, I cannot dispute what you are saying, in response to a given force they will oscillate with the same amplitude.
My little excel simulator effectively runs the 'drop' command from LFS, so the mass starts its motion at the springs natural length, ie x0 in your diagram. This means its actual vertical distance from its equilibrium point (ie. x1) varies depending on G or downforce, because the equilibrium point is lower. Hence is has a longer distance to accelerate before decelleration begins. Hence it shows a larger amplitude of oscillation.
But in operating conditions, when the initial displacement has already been taken up, this isnt relevant and so amplitude would remain the same.
Just goes to show that you were right to go with mathematical proof rather than iterative simulation
So I suppose the reason that aero cars tend to run with extremely high spring frequencies is not so much to control the amplitude of oscillations at a given speed, but rather to minimise the difference between displacements at different speeds. Ie. If your ride height changes by 100mm depnding on how fast you're going, it would play merry hell with your suspension geometry.
It is very much a case of our assumptions and initial positions being different. Again, I cannot dispute what you are saying, in response to a given force they will oscillate with the same amplitude.
My little excel simulator effectively runs the 'drop' command from LFS, so the mass starts its motion at the springs natural length, ie x0 in your diagram. This means its actual vertical distance from its equilibrium point (ie. x1) varies depending on G or downforce, because the equilibrium point is lower. Hence is has a longer distance to accelerate before decelleration begins. Hence it shows a larger amplitude of oscillation.
But in operating conditions, when the initial displacement has already been taken up, this isnt relevant and so amplitude would remain the same.
Just goes to show that you were right to go with mathematical proof rather than iterative simulation
So I suppose the reason that aero cars tend to run with extremely high spring frequencies is not so much to control the amplitude of oscillations at a given speed, but rather to minimise the difference between displacements at different speeds. Ie. If your ride height changes by 100mm depnding on how fast you're going, it would play merry hell with your suspension geometry.
iirc on cars that produce lots of downforce (formulae group-c prototypes) they use soft tyres with high sidewalls that part of the suspension setup rather than just being soft in order to grip the road ...
It looks like the argument was resolved, but I was going to say in a real-world application the amplitude stays the same, because increased movement due to downforce in one direction is exactly offset by decreased movement in the other direction, due to the same force.
Anyway, besides changing ride heights screwing up your suspension geometry, there's also the basic issue of keeping the car low to the ground--by using stiffer springs, you can have a lower ride height without bottoming out in the bumps and corners. Ignoring ride height regulations and track surface conditions, in the spirit of maximizing performance you'd probably keep going lower and stiffer until the frequency got so high the car was impossible to control.
Hehe
Hehe, that's not an equal systems to compare. See you drop your car from a bridge with 9,8 m/s^2 gravity. Firstly, you need to push your car to a top of the bridge by your hands, 'cause of broken engine. Yeah, there's no need to drop off a good working car, right?
I know you can do it, and we'll have a force A from the car hitting the ground (or water, milk, whatever you want))Now imagine you need to do the same, living in a world of 19,6 m/s^2 gravity. We've got a A*A force from hitting a car from same height. But as we see, you hardly managed half of the bridges height and drop it from there.
Because there is to type of energy - kinetic and potential. An object on same height would have more potential energy if the gravity is bigger. But this energy isn't appear from noting.
Maybe I don't understand the sprung-unsprung english terms but I don't know how to do that
I've tried to do some maths with some setups and sometimes I get reasonable results and sometimes I get stupid ones. I don't know where is my mistake.Could someone post an example of how to translate the spring stiffness to Frequency? take for example the Default setup of the XF GTi.
OK going with XF GTi hard track setup:
Front
stiffness: 50,000 N/m
mass of vehicle: 1026kg (including driver and 50% fuel)
unsprung mass: 50.6kg
sprung mass = total - unsprung = 975.4kg
front weight distribution: 58.2%
front sprung mass = sprung * dist = 567.7kg
mass on spring = front sprung / 2 = 283.8kg
frequency = 0.5 / ( PI * ( mass on spring / stiffness ) ^ 0.5 ) = 2.11 Hz
I'm too lazy to do the rear as well, but there's a worked example.
Front
stiffness: 50,000 N/m
mass of vehicle: 1026kg (including driver and 50% fuel)
unsprung mass: 50.6kg
sprung mass = total - unsprung = 975.4kg
front weight distribution: 58.2%
front sprung mass = sprung * dist = 567.7kg
mass on spring = front sprung / 2 = 283.8kg
frequency = 0.5 / ( PI * ( mass on spring / stiffness ) ^ 0.5 ) = 2.11 Hz
I'm too lazy to do the rear as well, but there's a worked example.
Great! That's just what I needed, thanks!! 
Nice one Bob. I've always seemed to find this confusing, but having seen it worked out it's all crystal clear again :P
*saves post into My Documents for reference later*
Quick Edit: I know where you get the total mass from (LFS duh!), but how d'ya know the unsprung masses, or is that you being clever with raf's again? Any chance you want to divulge the secrets and list the unsprung masses here for idiots like me :P
*saves post into My Documents for reference later*
Quick Edit: I know where you get the total mass from (LFS duh!), but how d'ya know the unsprung masses, or is that you being clever with raf's again? Any chance you want to divulge the secrets and list the unsprung masses here for idiots like me :P
Dunno about Bob (I assume he'll be along soon enough ) but I've calculated some of the unsprung masses using RAFs.
Basically, park the car on a flat place and leave it to settle, recording the fuel load. You can then export the 'Tyre Load' values from the RAF using F1PerfView. These seem to be spring loads rather than tyre loads so adding them (and dividing by 9.81) gives you the total sprung mass of the car. Subtract the fuel mass (using fuel tank size and density of fuel = 0.74 kg/litre) and you have the empty sprung mass. The difference between this and the LFS total mass is the unsprung mass
Here is what I've got so far...hope Bob can confirm this, otherwise one of us will look silly
XF GTi: 50.9kg
XR GT: 60.2
XR GT Turbo: 67.6
RB4 GT: 61.1
FXO Turbo: 64.6
LX4: 51.7
LX6: 57.7
Raceabout: 63.9
Formula XR: 59.7
Formula V8: 79.0
FXO GTR: 93.9
Hope that helps...I've not done all the cars yet!
) but I've calculated some of the unsprung masses using RAFs.Basically, park the car on a flat place and leave it to settle, recording the fuel load. You can then export the 'Tyre Load' values from the RAF using F1PerfView. These seem to be spring loads rather than tyre loads so adding them (and dividing by 9.81) gives you the total sprung mass of the car. Subtract the fuel mass (using fuel tank size and density of fuel = 0.74 kg/litre) and you have the empty sprung mass. The difference between this and the LFS total mass is the unsprung mass
Here is what I've got so far...hope Bob can confirm this, otherwise one of us will look silly
XF GTi: 50.9kg
XR GT: 60.2
XR GT Turbo: 67.6
RB4 GT: 61.1
FXO Turbo: 64.6
LX4: 51.7
LX6: 57.7
Raceabout: 63.9
Formula XR: 59.7
Formula V8: 79.0
FXO GTR: 93.9
Hope that helps...I've not done all the cars yet!
You're numbers don't quite match mine, but are close enough (within a kg). I think my numbers are slightly wrong anyway.
Completing the table...
FZ50: ..... 74kg
FZ50 GTR: 91kg
MRT5: .... 41kg
UF1000: .. 41kg
UF GTR: .. 41kg
XF GTR: .. 54kg
XR GTR: .. 94kg
Completing the table...
FZ50: ..... 74kg
FZ50 GTR: 91kg
MRT5: .... 41kg
UF1000: .. 41kg
UF GTR: .. 41kg
XF GTR: .. 54kg
XR GTR: .. 94kg
Fuel mass effect on suspension frequencies
(57 posts, started )
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