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| Expand | ||
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When the car brakes, it applies more force to the front wheels, thus less force to the back wheels. As the coilovers and shocks are under compression in the resting position, they will extend to meet the ground. In braking there is a backwards force applied to the wheel mounting point. This will create a clockwise moment around the pivot point, pushing the wheel down. This will contribute to an increase in downward force on the wheel. This will ideally distribute the load transfer under braking better than if it was below the horizontal. Realistically, the moment generated will be negligible (assuming small angles) to the rest of the load transfer seen by the loading condition, and therefore we will not adjust the mass proportion present on the back. |
Loading Cases From the Wheel (must pass all four loading cases)
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Case 1
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Case 2
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Case 3
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Case 4
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4355.64 N Upward
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4355.64 N Upward
...
4355.64 N Upward
...
4355.64 N Upward
...
2177.82 N Forward
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2177.82 N Forward
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1569.6 N Backward
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1569.6 N Backward
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2177.82 N Inward
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2177.82 N Outward
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2177.82 N Inward
...
2177.82 N Outward
...
| title | Reasoning |
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The regulations define the loading condition in terms of acceleration; 1G Steering, 1G Braking, 2G Bump (https://www.americansolarchallenge.org/ASC/wp-content/uploads/2021/12/ASC2024-Regs-EXTERNAL-RELEASE-A.pdf, Appendix F, F.2). We can see how load transfer effects the loading conditions on the wheels. Since the trailing arms are used on the rear of the car we’ll specifically look at those values.
...
The above screenshot is from a spreadsheet I developed that will calculate the the load distribution under a 1G brake and 1G steer. In a braking scenario there is more weight loaded to the front however. So I made some adjustments to the calculator and we can see that the rear should expect around 111 kg of mass.
...
So by the 2G bump case, we should expect an upward of force of around 2177.82 N
How the steering and braking cases impact our loading conditions are through the friction between the tire and the ground. If the car was turning with a centripetal acceleration of 1G, it would require a force equal to it’s weight. This force would be supplied by the friction from the tire, which is calculated by the coefficient of static friction and the normal force. Based on generally accepted values, the coefficient has a value less than 1, which means it cannot produce a force equal to the the weight of the car.
So now we need to make a decision. The car needs to be safe, but we don’t want unrealistic loading conditions either. Based on our priority to have a race-worthy car rather then a high performing car, we’ll assume the higher loading which means that the expected loading is:
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Loading Condition
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Load
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Notes
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2G Bump
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2177.82 N Upward
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Will be applied in all simulations
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1G Acceleration
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1088.91 N Forward
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Will create higher compressive stresses in the part
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1G Brake
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784.8 N Backward
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Will create tensile stresses in the part
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1G Steer Left
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1088.91 N Inward
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1G Steer Right
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1088.91 N Outward
There two additional to consider on top of this; the application of a safety factor, and the location of the loading.
At the time of writing this, the highest safety factor that needed to be applied in the previous car was 2 and therefore will be applied to this sprint. However, if this changes I’ll add it to the end of this section. So the new loading conditions are:
...
Loading Condition
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Load
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2G Bump
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4355.64 N Upward
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1G Acceleration
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2177.82 N Forward
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1G Brake
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1569.6 N Backward
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1G Steer Left
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2177.82 N Inward
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1G Steer Right
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2177.82 N Outward
By the regulations, the loading condition is applied to the contact path of the tire (where the tire makes contact with the road). Because of this separation between the application point and the wheel mounting point, there will also be moments generated around the wheel loading point. The unloaded diameter of the wheel is 557 mm, so, there will be a moment from the acceleration/brake condition and the steering conditions. The moment arm for the 2G bump case comes from the geometry of the wheel assembly WHICH NEEDS TBD. There is also pneumatic trail which would create more moments around the mounting point, which should not be neglected. Based on the tire specifications under Sources and some comparison to other rolling resistance coefficients, the rolling resistance coefficient given is likely not a unitless coefficient, but a measurement of the pneumatic trail as rolling resistance is the force required to overcome the moment created due to a non-uniform loading at the tire. Units are not provided, and thus will be assumed as millimeters. Meaning the pneumatic trail is 3.02 mm at it’s worst (larger moment arm, larger moment).
NEEDS A DECISION
At max compression the shock should be tangent to a circle concentric to the pivot point.
| Expand | ||
|---|---|---|
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Based on M = F*r*sin(theta) where theta is the angle between the moment arm, we see that if the angle is anything other than 90 degrees the moment generated is not maximized. At max compression, when the spring will need to apply the greatest force, losing any of it to a bad angle seems suboptimal. |
GET THE MOUNTING PATTERN FOR THE MOTORAt max compression the shock should be tangent to a circle concentric to the pivot point.
| Expand | ||
|---|---|---|
| ||
Based on M = F*r*sin(theta) where theta is the angle between the moment arm, we see that if the angle is anything other than 90 degrees the moment generated is not maximized. At max compression, when the spring will need to apply the greatest force, losing any of it to a bad angle seems suboptimal. |
GET THE MOUNTING PATTERN FOR THE MOTOR
Loading Cases From the Wheel (must pass all four loading cases)
Case 1 | Case 2 | Case 3 | Case 4 |
|---|---|---|---|
4355.64 N Upward | 4355.64 N Upward | 4355.64 N Upward | 4355.64 N Upward |
2177.82 N Forward | 2177.82 N Forward | 1569.6 N Backward | 1569.6 N Backward |
2177.82 N Inward | 2177.82 N Outward | 2177.82 N Inward | 2177.82 N Outward |
| Expand | ||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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The regulations define the loading condition in terms of acceleration; 1G Steering, 1G Braking, 2G Bump (https://www.americansolarchallenge.org/ASC/wp-content/uploads/2021/12/ASC2024-Regs-EXTERNAL-RELEASE-A.pdf, Appendix F, F.2). We can see how load transfer effects the loading conditions on the wheels. Since the trailing arms are used on the rear of the car we’ll specifically look at those values.  The above screenshot is from a spreadsheet I developed that will calculate the the load distribution under a 1G brake and 1G steer. In a braking scenario there is more weight loaded to the front however. So I made some adjustments to the calculator and we can see that the rear should expect around 111 kg of mass.  So by the 2G bump case, we should expect an upward of force of around 2177.82 N How the steering and braking cases impact our loading conditions are through the friction between the tire and the ground. If the car was turning with a centripetal acceleration of 1G, it would require a force equal to it’s weight. This force would be supplied by the friction from the tire, which is calculated by the coefficient of static friction and the normal force. Based on generally accepted values, the coefficient has a value less than 1, which means it cannot produce a force equal to the the weight of the car. So now we need to make a decision. The car needs to be safe, but we don’t want unrealistic loading conditions either. Based on our priority to have a race-worthy car rather then a high performing car, we’ll assume the higher loading which means that the expected loading is:
There two additional to consider on top of this; the application of a safety factor, and the location of the loading. At the time of writing this, the highest safety factor that needed to be applied in the previous car was 2 and therefore will be applied to this sprint. However, if this changes I’ll add it to the end of this section. So the new loading conditions are:
By the regulations, the loading condition is applied to the contact path of the tire (where the tire makes contact with the road). Because of this separation between the application point and the wheel mounting point, there will also be moments generated around the wheel loading point. The unloaded diameter of the wheel is 557 mm, so, there will be a moment from the acceleration/brake condition and the steering conditions. The moment arm for the 2G bump case comes from the geometry of the wheel assembly WHICH NEEDS TBD. There is also pneumatic trail which would create more moments around the mounting point, which should not be neglected. Based on the tire specifications under Sources and some comparison to other rolling resistance coefficients, the rolling resistance coefficient given is likely not a unitless coefficient, but a measurement of the pneumatic trail as rolling resistance is the force required to overcome the moment created due to a non-uniform loading at the tire. Units are not provided, and thus will be assumed as millimeters. Meaning the pneumatic trail is 3.02 mm at it’s worst (larger moment arm, larger moment). NEEDS A DECISION |
Timeline
Week 1 - Concepting
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