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Acceleration simulatorBest performance possible for vehicle acceleration
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| Stopping distance (100 − 0 km/h): 32 m Front/Rear brake force distribution: 70/30 |
Maximum slope angle (hill climbing): 47% (25°) |
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Speed unit for charts:
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Theory (more)
Theory
Weight transfer (more)
The major effect of the basic vehicle dimensions is weight transfer. Weight transfer is unavoidable. Although, with proper suspension design, it can be redistributed between two different wheels to maximize performance.
So, in most cases, the «perfect» vehicle would always have 25% of the total vertical force (weight and downforce) on each tire.
Lateral acceleration (more)
The maximum lateral acceleration (aLmax) can be found this way: (more)
Forces involved
Looking at the front view, the vertical forces acting on the vehicle are the weight (W = mg), pushing the vehicle down on the tires, and the aerodynamic lift (FL = 0.5ρCLAv2), lifting up the vehicle. Both are assumed to be equally distributed between the left and right tires and are acting one against each other.
Traction limit
The maximum inertia force (maLmax) that can be apply before the vehicle begins sliding sideways will depend on the friction force.
The resulting vertical force (Fv) acting on all tires is:
| Fv = (W − FL) | (1a) |
The resulting friction force (Ff) from all tires will be:
| Ff = μFv | (1b) |
That friction force will be the only force acting against the inertia of the vehicle, hence:
| maLmax = Ff = μFv = μ (W − FL) = μ (mg − 0.5ρCLAv2) |
Rewriting:
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(1c) |
Rollover limit
Because the friction force acts at ground level and the inertia force acts at the center of gravity, there will be a weight transfer from the inside tires to the outside tires. Once the vertical force acting on the inside tires is equal to 0, then the inside tires are off the ground, leaving the total weight of the vehicle on the outside tires. This is the rollover limit. The maximum load that can be transferred from the inside tires will be half of the total vertical force. Doing the sum of moments, we get:
| maLmaxh = 0.5 FvT = 0.5 (W − FL)T = 0.5 (mg − 0.5ρCLAv2)T |
Rewriting:
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(1d) |
Summing up equations (1c) and (1d):
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(1) |
Where K is defined by the smallest value of:
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And:
| ρ | = atmospheric air density (= 1.225 kg/m³) |
| CLA | = Lift factor |
| v | = vehicle speed |
| m | = vehicle mass |
| g | = gravitational acceleration (= 9.80665 m/s²) |
| μ | = tire-road friction coefficient |
| T | = vehicle track |
| h | = vehicle center of gravity height |
*Note: If aerodynamic effects are omitted (CLA = 0), then aLmax = Kg.
Longitudinal acceleration (more)
Forces involved (more)
In addition of the inertia force (ma) and the tire friction force (Ff = μFv), two other forces must be considered for evaluating longitudinal acceleration: the rolling resistance and the aerodynamic drag.
Rolling resistance (more)
The rolling resistance (Fr = frFv) is a force that acts in a way similar to the friction force. The only difference is that the rolling resistance direction depends on the rolling direction of the tire and the friction force direction depends on the torque application. So sometimes the rolling resistance acts against the friction force (acceleration) and sometimes it works with the friction force (braking).
Since the rolling resistance and the friction force are always in the same plane, you could say that the result is a tire with «net» friction coefficient (= μ ± fr) that is greater or smaller depending if the vehicle is accelerating or braking.
The rolling resistance is very small and can be omitted in most cases. The net effect is usually a variation of 1-2% of the original tire friction coefficient. It is more important in off-road situations where that variation can go up to 30% (tire on sand for example).
Aerodynamic forces
Drag (more)
The drag force (FD = 0.5ρCDAv2) is acting at height hD above the ground. Although unrelated, we can assume that the drag force height is equal to the center of gravity height (h). This will simplify the mathematical equations.
Lift (more)
A lift force (FL = 0.5ρCLAv2) is generally produced by a typical vehicle. It is relatively small though, so it can be ignored in most cases. However, with race cars, not only do we try to avoid lift but we actually create «negative» lift, or downforce, with the help of inverted wings or ground effects.
This force will be distributed between the front and rear axle according to the vehicle design. Since it is the reaction forces at each axle that are important, we will divide the total lift force into two: the front lift force (FLf = 0.5ρCLfAv2) and the rear lift force (FLr = 0.5ρCLrAv2), such that FL = FLf + FLr (or CL = CLf + CLr).
Braking (more)
From the previous figure (and assuming four identical tires), we can define the total friction force (Ff), the total rolling resistance (Fr) and the net vertical force (Fv):
| Ff = Fff + Ffr = μFv | |||||||||||||||
| Fr = Frf + Frr = frFv | |||||||||||||||
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Since practically all vehicles brake with both front and rear axles, the maximum force coming from the brake system will be the friction force from all tires (Ff).
Also, if we sum up the forces in the horizontal direction:
| ma − FD = Ff + Fr | (2a) |
Stopping distance (more)
From equation (2a) we get:(more)
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(2b) |
This defines the maximum possible deceleration of the vehicle at any given speed v. To simplify future notations, we will write equation (2b) the following way: a = Kt + Kav2.
Solving the equations of motion, we can find the stopping distance d necessary for decelerating from speed v to 0 with equation (3):
(more)Because the velocity will vary throughout the deceleration process and that the acceleration depends on the velocity, we have to consider the instantaneous velocity and acceleration as the vehicle decelerates. Assuming s represents position, u the velocity, a the acceleration and t the time, then we can define velocity and acceleration in their derivative form:
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Rearranging both equations to eliminate dt leads to:
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Solving the left-side equation:
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To solve the right-side equation, we must integrate by substitution (see an example), as the acceleration depends on the velocity:
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Rewriting the equation:
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Finally, by using the logarithmic identity:
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(3) |
Where:
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And:
| ρ | = atmospheric air density (= 1.225 kg/m³) |
| CDA | = Drag factor |
| CLA | = Lift factor |
| m | = vehicle mass |
| g | = gravitational acceleration (= 9.80665 m/s²) |
| μ | = tire-road friction coefficient |
| fr | = rolling resistance coefficient |
This is the equation that is used to estimate the tire friction coefficient μ with a given stopping distance d in equation (3) of the TIRE COEFFICIENT section. Equation (3) was rewritten to isolate μ assuming CLA = 0.
Axle load distribution (more)
When braking, there will be a weight transfer from the rear axle to the front axle. It will be important to know the exact amount when we will design our brake system such that we can distribute adequately the brake force between the front and the rear axle, because the maximum brake forces at each axle will be:
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If we sum up the forces in the vertical direction:
| Wf + Wr = mg − FLf − FLr = Fv |
or:
| Wr = Fv − Wf |
Dividing by Fv,
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(4a) |
To find the portion of the vertical force on the front axle: (more)
Doing the sum of moments:
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According to equation (2a), this is also true:
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Isolating Wf and dividing by Fv:
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In most cases, we can assume that FLf/FL ≈ lr/L and/or that FL is relatively small such that:
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(4b) |
Where:
| μ | = tire-road friction coefficient |
| fr | = rolling resistance coefficient |
The first component of equation (4b), lr/L, is the portion of the vehicle's weight on the front axle (at rest) and the other component of equation (4b) is the weight transferred from the rear axle to the front axle.
Note also that the portion of the vertical force on the front axle cannot exceed 100%. At this point, the vehicle is doing a «stoppie» as shown in the next figure.
Accelerating (more)
From the previous figure (and assuming four identical tires), we can define the total force from the tires (Ft), the total rolling resistance (Fr) and the net vertical force (Fv):
| Ft = Ftf + Ftr | |||||||||||||||
| Fr = Frf + Frr = frFv | |||||||||||||||
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When accelerating, the power can be transmitted to the road with the front axle (FWD), the rear axle (RWD) or both (AWD) depending on the automobile layout.
Also, if we sum up the forces in the horizontal direction:
| ma + FD = Ft − Fr | (5a) |
From equation (5a) we get:(more)
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(5b) |
This defines the acceleration of the vehicle at any speed v. To identify the maximum possible acceleration, we need to know the maximum tractive force, either based on power or traction availability.
Also, the maximum speed (vmax) that can reach a vehicle will happen when its acceleration will be zero (i.e. no more speed gain).
Power limited (more)
Maximum tractive force (more)
The total tire force (Ft), at a given speed v, can be found if the power available at the tires (Pt) is known:
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(6) |
Engine power (more)
The engine power (Pe) is the product of its torque (Te) and angular velocity (ωe):
| Pe = Teωe |
The wheel power (Pt) is the product of the wheel's torque (Tt) and angular velocity (ωt):
| Pt = Ttωt |
In any power transmission system there will be some friction losses in the bearings and seals, but also between gear teeth, belt and pulleys, chain links or even due to oil churning. The efficiency of a power transmission system (ηt) is defined as:
ηt =
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| Drive system | ηt |
| Gear set | 0.975 |
| Synchronous belt | 0.975 |
| V-belt | 0.965 |
| Roller chain | 0.925 |
If more than one drive system are used in series, you have to multiply their efficiencies together to obtain the overall efficiency of the drive system.
For example, most vehicle have two gear sets in the transmission plus one in the final drive, so ηt = (0.975)3 = 0.927.
In most power transmission there will be some slipping involved, particularly at the clutch and at the tire contact patch. The slip of a power transmission system (i) is defined as:
(1 − i) =
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Typical values of slip for road vehicles is usually assumed to be 2-5% (Theory of ground vehicles, 2nd ed., J.Y. Wong, 1993, p. 200).
Finally, all of the mechanical components will have to accelerate before the engine torque reaches the tire, so this inertia needs to be added with respect to the total mass of the vehicle (see mass factor).
So the net power available for accelerating the vehicle will be:
| Pt = Ttωt = (ηtTe)[(1 − i)ωe] = ηt(1 − i)Pe |
Which needs to be divided by the mass factor. Since the power is a limiting factor for acceleration only at higher speed, an average value of the 2nd gear and higher is used for the mass factor in this site (Typical value: 1.075).
Considering everything, this site assumes that − between the wheel power and the engine power − there is an 83% efficiency for a typical road vehicle and an 86% efficiency for one with a direct drive system, like the one shown in the next figure (top fuel dragster).
Furthermore, because most vehicles cannot keep their engine at maximum power due to the drive system design (see next figure), this site assumes that only 95% of the maximum power is available throughout the run (or what could be referred to as average power). And vehicles that can stay at maximum power (with a CVT), usually have a lower efficiency of the same order.
Top speed (more)
From equations (5b) and (6), we get: (more)
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Multiplying both sides by vmax:
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| v | 3 |
| max |
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This cubic function of vmax can be solved with Cardano's method, giving:
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(7) |
Where:
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| vmax= |
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Traction limited (more)
No matter how much power the vehicle can produce, the total tire force (Ft), at a given speed v, cannot exceed the friction force available at the tires (Ff).
Maximum tractive force
For an AWD:
The power is available at the front and rear axles, so:
| Ft = Ftf + Ftr = μ(Wf + Wr ) = μFv |
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(8a) |
For a RWD:
Since the power is only available at the rear axle, we need to know the vertical force acting on the rear axle: (more)
Doing the sum of moments:
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From equation (5a):
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(8b-1) |
The maximum tire force is the friction force from the rear axle:
| Ft = μWr |
With the help of equation (8b-1) we find:
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(8b) |
For a FWD:
Since the power is only available at the front axle, we need to know the vertical force acting on the front axle: (more)
Doing the sum of moments:
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From equation (5a):
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(8c-1) |
The maximum tire force is the friction force from the front axle:
| Ft = μWf |
With the help of equation (8c-1) we find:
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(8c) |
Simplifying equations (8a), (8b) and (8c):
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(8) |
Top speed
From equations (5b) and (8), we get:
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Isolating vmax:
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(9) |
Where:
For an AWD: |
K1 = 1 | K2 = 1 | |||||||||||||||||||||||
For a RWD: |
K1 =
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K2 =
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For a FWD: |
K1 =
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K2 =
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And:
| ρ | = atmospheric air density (= 1.225 kg/m³) | |||
| CDA | = Drag factor | |||
| CLA | = Lift factor | |||
| m | = vehicle mass | |||
| g | = gravitational acceleration (= 9.80665 m/s²) | |||
| μ | = tire-road friction coefficient | |||
| fr | = rolling resistance coefficient | |||
| = portion of the vehicle's weight on the front axle | |||
| = portion of the vehicle's weight on the rear axle | |||
| = CG-height-to-wheelbase ratio | |||
| = portion of the vehicle's aerodynamic lift on the front axle | |||
| = portion of the vehicle's aerodynamic lift on the rear axle |
Speed, distance & time (more)
Let's assume that we take an initial «snapshot» of a vehicle during its run. We record the time ti, the vehicle is going at a speed vi and has already traveled a distance di.
When the vehicle reaches the speed vf , we take a final «snapshot» and we want to know what will be the new time tf and the new distance df.
We already know three things about the vehicle during this portion of the run:
- the speed increase, Δv = vf − vi ;
- the average speed, vavg = 0.5( vi + vf );
- the average acceleration, aavg, calculated with equation (5b) and vavg.
Then, by definition:
tf = ti +
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| df = di + vavg ( tf − ti ) |
This site uses a speed increase Δv which equals one thousandth of the known top speed of the vehicle. For example, if the top speed is 300 km/h, the calculations are done every time the speed increases 0.3 km/h.
Hill climbing (more)
When a vehicle goes up a hill, the gravitational acceleration (g) stays vertical. Because it is a vector, this acceleration will be splitted into two smaller components: One that will push the vehicle down against the road (g
Note that we will also assume that the vehicle is at rest, trying to «hold on» with engine power, so we will neglect aerodynamic effects and rolling resistance.
The maximum slope that can climb a vehicle based on its maximum tractive capabilities is:
For an AWD: (more)
Equating the forces along the road plane:
| mgsinθ = Fff + Ffr |
| mgsinθ = μWf + μWr |
| mgsinθ = μ(Wf + Wr) |
| mgsinθ = μmgcosθ |
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| tanθ = μ | (10) |
For a RWD: (more)
Doing the sum of moments, we can find the portion of the weight on the rear axle:
| LWr = lfmgcosθ + hmgsinθ |
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We can rely only on the rear axle traction, so Fff = 0. Equating the forces along the road plane:
| mgsinθ = Ffr = μWr | ||||
sinθ = μ
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sinθ = μ
(
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1 − μ
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(11) |
For a FWD: (more)
Doing the sum of moments, we can find the portion of the weight on the front axle:
| LWf = lrmgcosθ − hmgsinθ |
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We can rely only on the front axle traction, so Ffr = 0. Equating the forces along the road plane:
| mgsinθ = Fff = μWf | ||||
sinθ = μ
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sinθ = μ
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1 + μ
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(12) |
Flip over limit
Regardless of the automobile layout, there is a slope where the vehicle will flip over, i.e. the normal load on the front axle will be zero. The flip over limit is: (more)
Doing the sum of moments, we can find the portion of the weight on the front axle:
| LWf = lrmgcosθ − hmgsinθ |
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Since Wf = 0 when the vehicle will begin flipping over, then:
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(13) |
Where:
| μ | = tire-road friction coefficient | |||
| = portion of the vehicle's weight on the front axle | |||
| = portion of the vehicle's weight on the rear axle | |||
| = CG-height-to-wheelbase ratio |
Axle load distribution (more)
It can be useful to know the forces acting on the front and rear axles. Not only it will tell you if you can do a wheelie or not, but it it can also be useful when designing the suspension for high speed conditions (especially with high downforce).
The load on the front axle ( Wf ) is: (more)
From equation (8c-1):
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(14) |
Where:
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And:
| ρ | = atmospheric air density (= 1.225 kg/m³) | |||
| CLA | = Lift factor | |||
| m | = vehicle mass | |||
| g | = gravitational acceleration (= 9.80665 m/s²) | |||
| fr | = rolling resistance coefficient | |||
| = portion of the vehicle's weight on the front axle | |||
| = CG-height-to-wheelbase ratio | |||
| = portion of the vehicle's aerodynamic lift on the front axle |
As for Ft , it is the smallest value found with equation (6) or equation (8).
Equation (14) cannot exceed 100% and has to be greater than, or equal to, 0%.
Equation (4a) from the braking section still applies to find the load on the rear axle ( Wr ):
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(4a) |
To find the loads with respect to the vehicle's weight (as plotted on the graph on this site), we do the following transformation:
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Similarly:
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