Jens made this calculator for MSXV, I uploaded a copy of the Excel file to google drive. This is what this page will be explaining: https://docs.google.com/spreadsheets/d/1kUrsEHLuv7FpMME8J_mQ9zWea7GSaVsi/edit?gid=1414416308#gid=1414416308

(currently this page is rough notes for a presentation, maybe will flesh out later - Jon, Nov 2024)

Center of Gravity (CoG)

The first page in the excel file attempts to calculate the center of gravity of the car based on coordinates from the arbitrary car origin from the chassis assembly. For us, we don’t really care about this part since we know the weight distribution of our actual car is roughly 75% rear, 25% front, 400kg total. (picture with real numbers is…. somewhere)

Loading Cases

This is the big thing. In Dynamics, we need to simulate that all our components can pass ‘worst-case load scenarios’. But, how do we find these worst-case loads? This comes from the harshest conditions are car is expected to experience in operation: 1G steer, 2G bump, and 1G brake.

• Note that ‘1G’ means the car is experiencing enough force to accelerate equal to 1x gravity (9.81 m/s^2).

• G steer means that the car is experiencing 1G force in the left or right directions (correlating to the centripetal force experienced during left or right turns)

• 2G bump means the car accelerates upwards at 2x 9.81m/s^2, which may only ever happen in very small time windows, namely if the car goes over a large bump

• 1G brake is if the car is slowing down at, you guessed it, 9.81m/s^2 (force is going backwards)

• There is also 0.5G acceleration, which is the opposite (force acts forwards). This is only 0.5G and not 1G because our motors will never ever get us to accelerate faster than 0.5G (this isn’t formula)

The fact that there are multiple different ‘worst-case scenarios’ for the car means that there are different loading cases we have to sim all our parts to!

MSXV’s Loading Cases:

The 5 loading cases (including ‘rest’) are shown as below. Note that for all 5 cases, we assume the car is undergoing 2G bump. So, the cases are actually 2G bump + whatever forward/back or left/right force is represented in that loading case

Open

Determining Loading Forces at Wheels

To sim parts, you need to know how much force they experience. The easiest ‘known’ source of force for a component comes from each wheel. But, we don’t currently know how much force each wheel experiences under each loading case. This is where the more complex math comes in:

Solving Weight Distribution

The Center of gravity of our car can be simplified with the following nodal diagram, taken from a side view:

Open

If the acceleration of the car and the distance of the CoG from both wheels is known, the fraction of the weight of the car supported by the front and rear wheels can be easily calculated with the formulas above.

This also holds true if a ‘front view’ of the car is taken. The left or right acceleration of the car will then dictate the left-vs-right weight distribution of the car.

Note that due to the T*G term above, the acceleration of the car affects the weight distribution. Thus, each loading case has their own weight distribution ratios.

The final result of these calculations will be a percent distribution between front-vs-rear, and another pair of fractional weight distributions for right-vs-left:

Weight at Each Wheel

Now, we need to know the actual weight on each of the 4 wheels, not just front vs rear and left vs right. Thankfully, this is quite simple. To find the weight ratio of a given WHEEL, just multiply the associated front/back ratio by the associated left/right ratio.

For example, for the ‘Accel + Right Turn’ case, the loading of the front-left wheel is 0.3344 * 0.5871

Now, this is just the wheel weight fraction. To the find the actual weight that each wheel supports, you just multiply that wheel fraction by the total mass of the car (assumed to be 300kg in the calculator, is actually ~400kg irl)

Forces at each Wheel

We’re almost done! Now that we know the weight of the car supported by each wheel (aka the ‘mass’ of each wheel) we just need to know how to convert that into a force (because forces are what cause our things to break and what we sim for). Thankfully, this is very easy because we can just use F = m*a !

That is, for a given wheel, if we know that we are experiencing 2G bump, the force in the y-direction (upwards) is equal to m * 9.81 * 2 , where 2 is from 2G, and m is the ‘mass’ number pulled from the orange table shown above.

Note the coordinate convention of the car (which is the same convention we will use for MS16) is as follows: X is to the left, Y is upwards, and Z is forward.

(Hold on a) Moment

One last thing to consider is that we like to imagine forces acting on the center of each wheel, because that makes life easy. However, reality is often disappointing….

The forces we solved for above actually apply at the bottom surface of the tire that touches the ground, known as the ‘contact patch’. So, in order to translate forces from the contact path to a ‘nicer’ location (like the center of the rim), we have to use basic r x f force translation (you’ll learn it in statics)

(Extra Note)

Technicallyyyy, for the purpose of simming suspension components, the moment about the y-axis can be ignored. This is because the brakes are what resist any moment about that axis (the axis of wheel rotation), and the hub or motor will not be resisting it. Therefore, the suspension, inner hub, etc will not resist that moment either

Nodal

Now that we finally know forces at each wheel, determining the loads for your specific dynamics parts design (primarilly suspension) requires even more math, usually based on the principles of statics. You’ll probably want Excel, Python or Matlab to solve this stuff for you. That is to say, you will need to do math to find out how forces from the contact path will translate and be resisted by the geometry of your parts!

Check out this in-depth page for the nodal and calcs we did for MSXV FSU: FSU Loading Calcs

(Sim Convention)

By convention, we choose to design around the left side of the car. That is, under a ‘1G left turn’ case, there will be more loading on thee left side, and vice-versa for the right side.

SUMMARY!!!

That was a lot, I don’t expect you to get everything. For new members, please at least remember the following:

1. There are 5 loading cases we care about (each with 2G bump as well):

   1. rest

   2. right turn + acceleration

   3. left turn + acceleration

   4. right turn + braking

   5. left turn + braking

2. For each loading case, we can find out what percent of the car’s total mass is sitting on each wheel

3. Knowing the mass on each wheel lets us determine how much force is applied on the wheel

4. Force is applied on the contact patch, NOT at the center of the wheel

5. We have to use Nodal and other math solvers to resolve the forces at the contact patch into the components we design and sim