How to use my data of Ixx, Iyy and Izz?

In summary, the conversation discussed the use of CAD and excel to determine the inertia momentum values for a Formula Student Vehicle, as well as the relationship between these values and the car's tendency to squat, dive, or spin. The specified question focused on how to involve the pitch inertia momentum in calculations for straight line acceleration, taking into account factors such as weight distribution and power transfer mechanisms. The conversation also mentioned the potential impact of inertial moments on the rotation and stability of the vehicle.
  • #1
LIL2606
9
1
Hi everyone, thank you for reading my thread I know the question is weird but I can't seem to explain it otherwise.

My problem:
Basically I am finishing a final year project which I based on a Formula Student Vehicle. Using CAD I have the vehicle modeled, and through the model analysis and with the help of excel I have a CG value for the whole car, and I have inertia momentum values for each sub assembly which I combined using the parallel axis theorem to get Ixx Iyy and Izz for the whole car. Roll Pitch Yaw.

General Questions:
Are they supposed to characterise the loads through the car? Pitch - Squat and Dive, Yaw - over steer under steer, Roll (??) - Cornering lateral loads?

After research:
The way I wrapped my head around this so far, and came up with the idea that it might have to do something with the load transfers, is that inertia is a characteristic describing a body's tendency to stay unchanged. These inertia momentum values are part of a matrix and this matrix describes the values of a rotating body in different positions. So the two combined and I might be wrong here, is that these values are describing how the car "doesn't want to" dive or squat or, spin... but I can't seem to find any evidence behind this, and while I know my whole car's Ixx Iyy and Izz values, I can't involve them in any of my calculations, when I look at for example on a straight line acceleration case..

Specified question:

If I tried to calculate a straight line acceleration - with a vehicle 45%FWD weight distribution, 1600 wheelbase, 200 kg, for a 1 meter fraction of the track, from velocity = 0, slip or power limited acceleration (I'd calculate both and would go with the smaller value), unequal length double A-arm suspension, with Prod and a spring. - how would I involve Iyy, the pitch inertia momentum which (decides) how much the car will (not want to) squat?

Thank you very much!
 
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  • #2
LIL2606 said:
Specified question:
If I tried to calculate a straight line acceleration - with a vehicle 45%FWD weight distribution, 1600 wheelbase, 200 kg, for a 1 meter fraction of the track, from velocity = 0, slip or power limited acceleration (I'd calculate both and would go with the smaller value), unequal length double A-arm suspension, with Prod and a spring. - how would I involve Iyy, the pitch inertia momentum which (decides) how much the car will (not want to) squat?
Welcome to the PF.

Others can give you better ME-centric replies, but it would seem to me that your 45%FWD weight distribution will change quickly (certainly within the first meter of acceleration) as the vehicle launches. The change in the weight distribution will affect the traction at the front and rear wheels, and combined with any power transfer mechanism (limited slip differential? Locked front and rear?) will change how quickly you can accelerate with the available traction. Can you address those issues so the MEs who click in next can help with better responses? Thanks.
 
  • #3
Its a Drexler Formula Sae differential - limited slip diff I believe its rear locking the front is independent.
 
  • #4
berkeman said:
Welcome to the PF.

Others can give you better ME-centric replies, but it would seem to me that your 45%FWD weight distribution will change quickly (certainly within the first meter of acceleration) as the vehicle launches. The change in the weight distribution will affect the traction at the front and rear wheels, and combined with any power transfer mechanism (limited slip differential? Locked front and rear?) will change how quickly you can accelerate with the available traction. Can you address those issues so the MEs who click in next can help with better responses? Thanks.
Thank you very much for your reply! Ah I knew I forgot some details.. thanks for helping by pinpointing them!
 
  • #5
Those inertial moments have nothing to do with linear acceleration except indirectly. One can use them to calculate the rotation of the vehicle in response to rotational forces. Then the orientation of the vehicle may change your calculations of linear acceleration.

For instance, consider that the rear tires contact point with the ground are below the vehicle center of gravity (CG). So there is a rotational force that wants the vehicle to do a "wheelie". You can use the inertial moment to determine how much the front end lifts up. If this was a motorcycle, the wheelie might easily be so large that it flips over and crashes instead of accelerating down the road. I don't know if that might happen in a stock car, but I'm sure it could happen to a dragster.
 
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  • #6
FactChecker said:
Those inertial moments have nothing to do with linear acceleration except indirectly. One can use them to calculate the rotation of the vehicle in response to rotational forces. Then the orientation of the vehicle may change your calculations of linear acceleration.

For instance, consider that the rear tires contact point with the ground are below the vehicle center of gravity (CG). So there is a rotational force that wants the vehicle to do a "wheelie". You can use the inertial moment to determine how much the front end lifts up. If this was a motorcycle, the wheelie might easily be so large that it flips over and crashes instead of accelerating down the road. I don't know if that might happen in a stock car, but I'm sure it could happen to a dragster.
Yes, what you described is called squat for formula cars. The nose slightly lifts up, the weight shifts backwards and the rear of the car sinks towards the asphalt, so the whole car "rotates" around its Y axis, and in my understanding the inertia momentums Iyy in this case, tells how much the car "resists" this rotation?

But yeah you are right I need to rephrase my question a bit, I think I am trying to find the "state" of the car in every metre, with respect to time.
So at distance =0 velocity =0 time =0
the car is at rest WD = 45% Fwd etc.

At distance = 1m velocity =? t =? WD =Fwd? and so on?
 
  • #7
Then you are heading in the right direction. After calculating the force of the tire on the road surface, you can calculate the horizontal linear acceleration of the CG using F = mA. As a separate calculation, you can calculate the rotational rate acceleration, ##a##, of the car using the resulting torque, ##\tau## and the moment of inertia, ##I_{yy}##: $$\tau = I_{yy} a$$

PS. To do an accurate calculation of the motion, you need to calculate the motions as time steps along. Rotating the car lifts the CG and gets gravity involved. The entire problem in general uses the 6 Degree Of Freedom (6-DOF) Equations Of Motion (EOM). (see https://en.wikipedia.org/wiki/Six_degrees_of_freedom ) that are often used in airplane simulations. In you case, you may be able to ignore some of the rotations and motions and greatly simplify the simulation. There are tools in simulation software that can help you to do the full-complexity simulation. (for instance, see MATLAB/Simulink, https://www.mathworks.com /help/aeroblks/6dof.html ).
 
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  • #8
FactChecker said:
Then you are heading in the right direction. After calculating the force of the tire on the road surface, you can calculate the horizontal linear acceleration of the CG using F = mA. As a separate calculation, you can calculate the rotational rate acceleration, ##a##, of the car using the resulting torque, ##\tau## and the moment of inertia, ##I_{yy}##: $$\tau = I_{yy} a$$

PS. To do an accurate calculation of the motion, you need to calculate the motions as time steps along. Rotating the car lifts the CG and gets gravity involved. The entire problem in general uses the 6 Degree Of Freedom (6-DOF) Equations Of Motion (EOM). (see https://en.wikipedia.org/wiki/Six_degrees_of_freedom ) that are often used in airplane simulations. In you case, you may be able to ignore some of the rotations and motions and greatly simplify the simulation. There are tools in simulation software that can help you to do the full-complexity simulation. (for instance, see MATLAB/Simulink, https://www.mathworks.com /help/aeroblks/6dof.html ).
Yeah I think that's what I am looking for, the Equations of Motion where we have 6 degrees of freedom, or are they not put into stone like Newton I or Newton II? I am currently looking at straight line acceleration, but I would like to involve cormering, which will involve the other 4 degrees as well.

The aim is to calculate the state of motion / the state of the car in small increments it can be seconds doesn't have to be metres, only I will need to know how far the car traveled in that 1 second.

So Iyy will not be a magnitude it will be a rate of how quickly how much weight will transfer, and I'm guessing it will be an increasing number.. so the static case I have calculated will be my minimum? Or that will be the maximum number before the car does a wheelie?

Now to further confuse myself, I think only the Sprung mass will be affected by the squat, so I will have 2 torques that are opposing each other as the sprung mass will rotate (side view - car travels from right to left) clockwise but the unsprung mass will try to move the car forward so it will try to rotate it anti clockwise , and when they are equal, my wheels will spin. Is my logic correct? If so how do I write this down?

I am quite alright with theoretical physics but horrible in applied physics...
 
  • #9
LIL2606 said:
Yeah I think that's what I am looking for, the Equations of Motion where we have 6 degrees of freedom, or are they not put into stone like Newton I or Newton II? I am currently looking at straight line acceleration, but I would like to involve cormering, which will involve the other 4 degrees as well.
Do you want a real simulation of that based on valid physics or just one that looks reasonable to a person playing a video game? The real physics is a difficult thing to simulate correctly. The spring forces are sensitive to small amounts of compression and the tire behavior depends a lot on specific tread charactoristics. I am not familiar with how it is done in a video game and I have never done much with ground handling of a vehicle. So I can not help much. But I will give you some of my thoughts on simulating the true physics of this problem, even if they are not very authoritative.
The aim is to calculate the state of motion / the state of the car in small increments it can be seconds doesn't have to be metres, only I will need to know how far the car traveled in that 1 second.
This will require much smaller time steps. I would expect at least one thousand time steps (aka "frames") per second.
So Iyy will not be a magnitude it will be a rate of how quickly how much weight will transfer, and I'm guessing it will be an increasing number.. so the static case I have calculated will be my minimum? Or that will be the maximum number before the car does a wheelie?
No. ##I_{yy}## is defined by the position of the mass relative to the CG. The things that would change it are the spring compression changing the relative position of the wheels and the fuel sloshing around in the gas tank. (and maybe other changes that I can not think of right now.)
Now to further confuse myself, I think only the Sprung mass will be affected by the squat, so I will have 2 torques that are opposing each other as the sprung mass will rotate (side view - car travels from right to left) clockwise but the unsprung mass will try to move the car forward so it will try to rotate it anti clockwise , and when they are equal, my wheels will spin. Is my logic correct? If so how do I write this down?
This doesn't sound right to me. The tires move in the same direction as the sprung mass of the car, but maybe at a different rate.
I am quite alright with theoretical physics but horrible in applied physics...
I think this will be a tough simulation to accomplish for anyone.
 
  • #10
It is a motorsports degree, so I would like to stick to physics not video games. Before I make it (more) complicated I am trying to make it work. If I can calculate it for a second it should be possible to break it down to smaller fractions later, but to be honest my project is not completely about the simulation, so braking it down to smaller fractions and refineing it.. someone else can do that next year haha!

I am pretty certain about the sprung and unsprung mass thing.. think about it! The body of the car wants to "flip up into the air" backwards, opposite rotational direction than the rotation of the wheels... but the suspension and what's connected to the wheels, are rotating the car "into the ground"

Iyy is defined by the position of the mass relative to the CG, but as the car accelerates and the mass shift backwards and the WD shifts back, technically the CG is shifting backwards? Increasing mass will result in an increasing Iyy?
 
  • #11
LIL2606 said:
It is a motorsports degree, so I would like to stick to physics not video games. Before I make it (more) complicated I am trying to make it work. If I can calculate it for a second it should be possible to break it down to smaller fractions later, but to be honest my project is not completely about the simulation, so braking it down to smaller fractions and refineing it.. someone else can do that next year haha!
In order to stop the simulation from going wild, very high frame rates may be necessary. In a one-second simulated frame of motion, gravity would move the car under the ficticious, simulated ground level and the spring compression force would be at its maximum. The next frame calculation would then shoot the car up into the air. The result would be that the car, even before it moves forward, would bounce up and down a lot.
I am pretty certain about the sprung and unsprung mass thing.. think about it! The body of the car wants to "flip up into the air" backwards, opposite rotational direction than the rotation of the wheels... but the suspension and what's connected to the wheels, are rotating the car "into the ground"
That doesn't sound right to me, but I am not sure what you mean.
Iyy is defined by the position of the mass relative to the CG, but as the car accelerates and the mass shift backwards and the WD shifts back, technically the CG is shifting backwards?
If you are talking about the gas in the tank shifting backward, that is correct.
Increasing mass will result in an increasing Iyy?
Right, unless the mass is also moved closer to the CG. Usually, mass is not increased during the simulation; as gas is burned, the mass will decrease.

PS. I recommend that all your simulation code use a frame time step that is a top-level parameter so you can easily change it. Different situations and simulation models may require different simulation time steps.
 
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  • #12
I'm not trying to simulate as such! All I want to do is enter formulas into excel and get it to spit some numbers out for me that I can explain. If I am really fancy, then plot the numbers on some graphs. No simulated ground or bouncing should take place, but let's not get stuck on the increments anyway I can make my excel sheet 10.000 lines long for 5m travel I just don't think that it will be necessary.

Sprung unsprung, think of a motorbike wheelie, the front tire comes up in the air the rider's back will hit the ground if he is not careful but the wheel that is touching the ground is rotating forwards, the opposite direction right?
 
  • #13
Ok. That makes sense. The reason that the ground is "simulated" is to calculate how much spring compression there is and the amount of force with which each wheel presses on the ground. It can be as simple as the height=0 plane.
 
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  • #14
Okay so in the moment the car launches F1=mass.car*a force acts on the tire contact patches and the full mass of the car F originates from the engine power and torque through the final drive system. The sprung mass of the car starts a different rotation, the force acting on it is the same(?) but opposite(?)
f2=mass.??*r*alpha which from we can calculate r if we assume the car sits all the way down as much as it can, and the springs are completely compressed. We can then calculate T=m*r^2*alpha from which m*r^2 is Iyy and that will give the maximum Iyy. It is the maximum because the springs won't allow for any larger angle. In these equations the r basically describes the load transfer as well right?

So the net force acting on the car at this moment, when the car launches, will be F1-F2=Fnet this Fnet= m*a which gives the a, as for linear forces along the X axis rotational forces along the Y axis and I need to add the traction limit for this acceleration, that's for the normal forces along Z acting. From the traction limit we can calculate the load transfer. Rotation around X and Z are assumed to be 0 and translation around Y and Z are also assumed to be 0. That describes all 6 degrees of freedom right?

So I can fraction it by the spring compression.. assuming the springs compress to 3/4 of their length first, then half then 1/4 then fully compressed, or - More reasonably, I should be able to fraction it by the engine output, and somehow determine the spring compression from the weight transfer. F2=m*r*alpha=-kx (for compression) and it makes sense, as when r is 0 there is no weight transfer, but when r is the largest k is a constant so x has to be at its smallest. Right?

What do you think?
 
  • #15
Here is the general outline of the calculations that you could do in a simulation:
1) On one time frame, you know the exact position, orientation, and velocities of the vehicle, wheels, etc.
2) From that, you can calculate the force from spring compressions for each wheel (and therefore, the contact pressure between each tire and the ground.)
3) Knowing all the forces, gravity, position, velocities, etc., you can use the 6-DOF EQMO to determine the motion for a tiny bit of time where those known values remain close enough. That gives the resulting positions, velocities, orientation, for the next frame.
4) Go to step 1 and repeat.

I have no experience with any other method and can not really help you with the approach that you seem to be describing.
 
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  • #16
FactChecker said:
Here is the general outline of the calculations that you could do in a simulation:
1) On one time frame, you know the exact position, orientation, and velocities of the vehicle, wheels, etc.
2) From that, you can calculate the force from spring compressions for each wheel (and therefore, the contact pressure between each tire and the ground.)
3) Knowing all the forces, gravity, position, velocities, etc., you can use the 6-DOF EQMO to determine the motion for a tiny bit of time where those known values remain close enough. That gives the resulting positions, velocities, orientation, for the next frame.
4) Go to step 1 and repeat.

I have no experience with any other method and can not really help you with the approach that you seem to be describing.
Can you please write down the 6DOF EQOM-s for me please, and or give me a link where I can find them?
 
  • #17
LIL2606 said:
Can you please write down the 6DOF EQOM-s for me please, and or give me a link where I can find them?
I think that you would be better off using Google and picking one that matches your notation and knowledge level to your satisfaction. Then if you have specific questions on that, you can post the link and ask your questions.

PS. I recommend that you start "simple". Don't worry about sprung versus unsprung weight or about the different rotation of the wheels. Getting all that correct is a large job. At most, start with a single mass, and mass properties, and the spring compression with its corresponding force and damping. That may be hard enough.
 
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1. What is the purpose of using data of Ixx, Iyy, and Izz?

The data of Ixx, Iyy, and Izz are used to calculate the moments of inertia of an object. This information is crucial for understanding an object's rotational motion and stability.


2. How do I find the values for Ixx, Iyy, and Izz?

The values for Ixx, Iyy, and Izz can be found by conducting experiments and measurements on the object or by using computer simulations. These values can also be provided in technical specifications for standardized objects.


3. Can I use data of Ixx, Iyy, and Izz for any shape or object?

Yes, the data of Ixx, Iyy, and Izz can be used for any shape or object as long as the object is rigid and has a defined axis of rotation.


4. How can I use the data of Ixx, Iyy, and Izz to improve my experiment or design?

The data of Ixx, Iyy, and Izz can be used to predict the rotational behavior of an object and identify potential areas of weakness or instability. This information can be used to make improvements and adjustments to the experiment or design.


5. Are there any limitations to using data of Ixx, Iyy, and Izz?

One limitation is that the data of Ixx, Iyy, and Izz are only applicable for rigid bodies. If the object has deformable or flexible components, the data may not accurately represent the object's rotational behavior.

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