Cornering

[To understand the material of these documents, I assume the reader have a firm understanding of Newtonian Kinematics.]

In most racing events, you need to make turns. This is where your driving skill kicks in. Everyone knows how to push the gas pedal to accelerate the car. But few knows how to attack a corner. In this page, I will try to find the best way to attack a 90 degree corner. You should be able to shed some lights from this analysis to apply the knowledge to other corners.

Traction Limit

Route A Time

v₂ = sqrt(0.94×9.8×29.3) = 16.429ms^-1

s₂ = π×29.3/2 = 46.023m

t₂ = s₂/v₂ = 2.801s

Next, we calculate the time we need to finish the exit. To do this, we need to solve the following quadratic equation:

50 = 16.429t₃ + 0.5×0.94×9.8t₃²

4.802t₃² + 16.429t₃ - 50 = 0

t₃ = 1.942s or -5.363s (rejected)

Then we need to know how long it takes to enter. First, we need to compute the maximum initial speed.

16.429²-u² = 2×-0.94×9.8×50

u = 34.427ms^-1

Then use it to comupte the travel time.

50 = 34.427t₁ - 0.5×0.94×9.8t₁²

4.802t₁² - 34.427t₁ + 50 = 0

t₁ = 2.023s or 5.146s (rejected)

Hence, the total time is:

t = t₁ + t₂ + t₃ = 6.766s

Route B Time

Using exactly the same approach as route A except for the value of the radius, we can get the Route B time.

v₂ = sqrt(0.94×9.8×50) = 21.462ms^-1

s₂ = π×50/2 = 78.538m

t₂ = s₂/v₂ = 3.659s

Next, we calculate the time we need to finish the exit. To do this, we need to solve the following quadratic equation:

50 = 21.462t₃ + 0.5×0.94×9.8t₃²

4.802t₃² + 21.462t₃ - 50 = 0

t₃ = 1.69s or -6.16s (rejected)

Then we need to know how long it takes to enter. First, we need to compute the maximum initial speed.

21.462²-u² = 2×-0.94×9.8×50

u = 37.173ms^-1

Then use it to comupte the travel time.

50 = 37.173t₁ - 0.5×0.94×9.8t₁²

4.802t₁² - 37.173t₁ + 50 = 0

t₁ = 1.733s or 6.008s (rejected)

Hence, the total time is:

t = t₁ + t₂ + t₃ = 7.082s

Route C Time

v = sqrt(0.94×9.8×100) = 30.35ms^-1

s = π×100/2 = 157.075m

t = s/v = 5.175s

Conclusion

It might seem that Route C is the best route among the three. In reality, this may or may not be the case. If the track after this is a long straight to the finish line, Route B might be the best bet. This is because the exiting speed of Route B is the highest. (37.173ms^-1 versus 34.427ms^-1 in Route A and 30.35ms^-1 in Route C) However, if there is not much straight ahead, Route A maybe better because the faster exit speed in Route B may not be able to compensate the time it lost for traveling a longer arc. analyze show you some guidelines to For those uninitiated, drag racing is a very simple racing sport. Two cars race a quarter mile straight line and see who gets to the destination first. So basically, it is a test of starting and acceleration techniques. Starting technique probably is not that useful in lap races, understanding how to obtain maximum acceleration is definitely important in all kinds of racing endeavors.

Drop the clutch

CAUTION: Doing this is bad for the transmission of your car!

When you set your gear at neutral, you can still push your gas pedal to rev your engine. Note that the torque generated by the engine is different at different engine speed. It is possible that if you engage the gear at a different engine speed, you get different torque to drive the wheels. Obviously, the best spot to engage (drop/release the clutch) the gear is at the peak torque RPM. However, since every car has a maximum acceleration (See the Maximum Acceleration/Deceleraton section), if the acceleration you get at peak torque exceeds that limit, your wheel will spin in the air and results in no acceleration. Therefore, you might need to do trial and error to find out which RPM can launch your car at maximum acceleration.

Power Shifting

CAUTION: Doing this is very very bad the transmission for your car! As an extension of clutch dropping, you push the gas pedal even when you are shifting in between the gears such that when you engage the next gear, you get more torque than you normally get.

Optimal Shift Point

If the torque curve of your car is peaky enough that it is better to shift up the car before the engine red line, you need to know the optimal shift points at each gear to extract the maximum acceleration from your car. For information about how to calculate optimal shift points, you can refer to the Optimal Shift Point section.

Autocrossing Techniques

Head and Toe Downshift

Drifting

When people talk about how fast a car can go, they usually quote the 0-100km/h time. But how do they measure it? In a manual car, in addition to putting your foot firmly on the gas pedal, you also need to shift the gear. But what is the best way to shift gear such that you can attain the top speed of your car in the shortest possible time?

In the last section, you learn that lower gear (higher gear ratio) gives you better acceleration. So you might well push the gas pedal until the engine red line and then shift up to the next gear. This is usually true for most cars. But if the torque curve of your car falls steeply after the torque peak, it is possible that you get more torque at the next gear. In that case, if you want to get the best acceleration, you need to find the optimal point to shift.

Simplified Torque Curve

Our Net Force formula tells us acceleration depends on the torque curve. Unfortunately, it is very hard to obtain precisely the whole torque curve of an engine through normal source. The only information we usually have are the peak power and peak torque. It turns out if we make some assumptions about the torque curve, these two numbers are enough for us to sketch one.

To simplify the calculations and get the most from the few data points we have, I assume the torque curve to be linear, ie it goes up linearly from 0 until the peak torque and then goes down linearly until the red line. Next, I use the power torque conversion formula to translate the peak power into a point on the torque curve after the torque peak. We further assume that the torque we have when the engine is at rest is 0. Remember that two points make a line? We can just connect the peak torque point with the 0 point and then connect it with the point we get from power torque conversion. Now we get a torque curve that goes up linearly from 0 to the peak torque and then from there it falls until it reaches the engine red line.

Our assumptions are not as weak as we might think, our Skyline's official torque curve isn't much different from a linear one.

Now it's the time to play with the data of our Skyline - 206000W@6800rpm, 392Nm@4400rpm. First, we need to know the slope of the line before the peak torque. That will be 392/(4400/60)=5.345. Next we need to get the torque curve point corresponding to the peak power. From the conversion formula the torque at that point is 206000/(2π×6800/60)=289.288Nm. We can then use this data point to find the torque curve after peak torque. The slope of the line after peak torque is 60×(288.288-392)/(6800-4400)=-2.568. The y-intercept of the line after peak torque is 392+(-2.568)×(-4400/60)=580.32. So the torque curve is:

P.S. If you insist in getting the real torque curve of your car, you can take it to a car tuning company to run a dyno test on your car.

Transmission Torque

Remember our gear-shifting strategy aims at maximizing the acceleration such that we get to the top speed in the shortest possible time. This is equivalent to maximizing the force generated by the engine. Our Net Force Formula told us the force generated by the engine is equal to torque times final drive ratio times k-th gear ratio divided by tire radius. Since we have the same tire radius and final drive ratio for each gear, it turns out that the only thing we need to care about is the product of torque and the k-th gear ratio. For the sake of convenience, we call this product, transmission torque.

To see whether an optimal shift point exists before the red line, we can calculate the transmission torque at red line and then compare it with the peak transmission torque in the next gear. If it is substantially lower, there is a good chance that there exists an optimal shift point before red line.

Gear	1st	2nd	3rd	4th	5th	6th	reverse
Gear Ratio	3.827	2.36	1.685	1.312	1	0.793	3.28
Peak Transmission Torque (Nm)	1500.18	925.12	660.52	514.3	392	310.86	1285.76
Transmission Torque at Red Line (Nm)	828.67	511.01	364.86	284.09	216.53	171.71	710.22

Our Skyline shows some sign of optimal shift point from the 1st to the 5th gear. To confirm that, we need to plot the transmission torque curves for each gear. To do this, we need the concept of transmission engine speed. Note that when you shift up a gear, the ratio between the current engine speed and the engine speed at the next gear equals to the ratio of the current gear ratio to the next gear ratio. So the 1000rpm in the 5th gear means 780rpm in the 6th gear for our Skyline. To put the engine speed in different gears on the same ground, we need to divide engine speed in each gear by its gear ratio. This new engine speed is called transmission engine speed.

As we can see there are intersections between the transmission torque curves, this confirms the existence of optimal shift points before red line.

Optimal Shift Points

Note that to maxmize acceleration, we need to shift such that the combined area below all these transmission torque curves is the largest. The only way to maximize it is to shift at the intersection points. From the graph, we see that the lines intersect at the line after peak torque happens. So by equating the transmission torques after peak torque in two consecutive gears, we can solve for the optimal shift point.

With this formula and convert the engine speed to RPM, we obtain the optimal shift point table.

Gear	1st	2nd	3rd	4th	5th	6th
Gear Ratio	3.827	2.36	1.685	1.312	1	0.793
Optimal Shift Point (RPM)	8387.29	7911.1	7623.54	7694.66	7562.64
RPM at next gear	5172.2	5648.39	5935.95	5864.83	5997.03
Transmission Torque at Optimal Shift Point (Nm)	847.13	570.49	428.06	329.31	256.66

Alternative Solution: Using Power curve to find optimal shift points

A further understanding of the relationship between Power and Torque reveals that the shift point calculations can be greatly simplified by looking at the Power curve.

Recall the formula P=Fv. Moving around the terms, you get a=P/(mv). This means that given the velocity of the car, the mass of the car and the power transmitted from the engine to the tires, the acceleration can be calculated from a=P/(mv). This means that the higher the car velocity is, the less acceleration you can expect.

Note that the power curve of a car can be obtained by P=2πΓ(ω)ω. If we take our Skyline's torque curve after the peak torque, we can obtain a power curve like P=2πω(-2.568ω+580.32). If you plot this power curve, then you know that it peaks at a certain point and then falls all the way from there. (Officially the peak power point is at 6800RPM but our peak based on our simplified torque curve most likely won't yield the same power curve and hence the same peak power point.)

From the power curve, we know that if we want to maximize the acceleration of the car, we don't need to up shift before the peak power point. This is because if we shift up, we get lower RPM and hence less power from the engine. Note that the velocity of a car before and after the shift remains the same. However, the acceleration of the car is greatly reduced because power is smaller at lower engine speed.

It is easy to see that to maximize acceleration, we need to keep the power from dropping. That means we should shift when we are after the power peak and if we don't shift at the next instant, we will find that we get less power from the engin by avoiding the up shift. That happens when we get the same power from the engine when we shift up. Suppose P is the power at the optimal shift point. By solving the quadratic equation: 2π(-2.568ω²+580.32ω)-P=0, the roots we obtain are the optimal shift point (ω) and the engine speed at the next gear after the shift (ωg^k+1/g^k). Note that the sum of the roots equals the negation of the second term divided by the first term, so we have the following equality:

Solve for ω, we get exactly the same equation we have in our previous discussion.

Conclusion

After all these discussions, you may want to ask: why there are optimal shift points in the first place? If the gearing designer did a good job, he/she should make the red line the optimal shift point in every gear. The answer to this question is that for most cars, the optimal shift point for every gear is on the red line. However, the power of this Skyline is undervalued due to a Japanese industry consensus that road cars can only have 280 horsepower. (ie 206000W) [On the other hand, the Germans electronically limit their cars' top speed to 250km/h] In reality, this consensus is only followed on paper.

Car engine modifications without a corresponding tuning in gearing is another possible cause that the optimal shift point is not on the red line. In this case, you either have to tune the gearing or if you are short on money, learn how to shift before red line. There will be a page talking about how to do gearing tuning. So keep reading! ;^)