In this exercise it is proposed some aerodynamic parameters from a fictitious car. The objective is to modify the adjustment parameters to reach the target ones. The vehicle baseline configurations is summarized below:

Where C_X∙S is the drag coefficient given in m², C_ZF∙S and C_ZR∙S are the downforce at front and rear axles, respectively. Another important data are the ones from the wings, since this problem approach an open-wheel race car there are two wings, thus two adjustment parameters, which are α and β that refers to the front and rear wing angle of attack, respectively.

On Figure 1 it is possible to extract useful information for this exercise. For instance, at the graph ΔC_X∙S × FW angle there is the slope of the line, which is given by 0.0044. This means that at each 1° variation of FW, C_X∙S, C_ZF∙S and C_ZR∙S vary by 0.0044, – 0.04564 and 0.00068, respectively.

On Figure 2 there are the information about the rear wing. It is possible to notice that for 1° angle variation, C_X∙S, C_ZF∙S and C_ZR∙S vary by 0.01005, 0.00082 and – 0.02313, respectively. With those data it is possible to understand the differences between the front and the rear wing. FW provides more gain in downforce without a high drag value relative to RW as can be seen below:

Eff_RW = C_Z∙S/C_X∙S = (0.00082 – 0.02313)/0.01005 = – 2.2199 ≃ – 2.2 → Rear wing efficiency

Eff_FW = C_Z∙S/C_X∙S (-0.04564 + 0.00068)/0.0044 = – 10.21818 ≃ – 10.2 → Front wing efficiency

As can be seen, the efficiency of the front wing Eff_FW is about 4.6 times higher than RW one.

Question 1 – Re-balance using the front wing

Given a starting configuration and the front wing polar, which is the required front wing angle for a front balance F_bal = 42% ?

	CX∙S	CZF∙S	CZR∙S	CZT∙S	Fbal	α	β
Base	1.330	– 2.280	-2.390	–	–	–	–

The first step is to calculate the total downforce C_ZT∙S and the front balance F_bal for the baseline configuration.

C_ZT∙S = -2.280 + (- 2.390) = – 4.670 m²

F_bal = – 2.280 / – 4.670 = 0.488222 ≃ 48.8%

	CX∙S	CZF∙S	CZR∙S	CZT∙S	Fbal	α	β
Base	1.330	– 2.280	-2.390	– 4.670	48.8	–	–

Since the objective is to reduce F_bal to 42%, the delta will be of 6.8%. This means that the vehicle balance is shifted towards the rear axle. Using the data sheet illustrated by Figure 1, it is possible to understand that this adjustment can only be done by reducing the front wing angle. Hence, FW characteristics at – 15° is summarized at Table 3.

Wing / Parameters	ΔCX∙S [m²]	ΔCZF∙S [m²]	ΔCZR∙S [m²]
Front wing @ 15°	– 0.067	0.668	– 0.010

The procedure is to sum the data described on Table 3 into the ones at Table 2.

	CX∙S	CZF∙S	CZR∙S	CZT∙S	Fbal	α	β
Base	1.330	– 2.280	-2.390	– 4.670	48.8	–	–
FW 15°	1.263	-1.612	-2.400	– 4.012	40.2	– 15	–

Hence, by the same procedure adopted to calculate the baseline F_bal, it was calculated for this new condition. Table 4 shows that the new front balance is below the target F_bal, which is 42%, thus a difference of 1.8%. The interesting point about this table is that was required 15° to reduce F_bal in 8.6%. This allows to find another quantity, ΔF_bal/Δα:

ΔF_bal/Δα = (40.2 – 48.8)/(- 15 – 0) = 0.5733 ≃ 0.6 [%/°]

This is an important quantity since the objective is to reduce F_bal from 48.8% to 42%, ΔF_bal from the baseline to the target is – 6.8%. Once it is calculated how much percent of F_bal the front wing is capable to produce for 1°, it is possible to calculate the exact angle to reach this F_bal.

ΔF_bal/Δα = 0.573 → Δα = ΔF_bal/0.573 = – 6.8/0.573 = 11.87° ≃ 12° → Δα = – 12°

As can be seen, it is requested just -12° to reach the 42% target F_bal. Now, using again the data illustrated at Figure 1 it is possible to calculate the deltas ΔC_X∙S, ΔC_ZF∙S and ΔC_ZR∙S since the graphs exhibit the slopes, in other words, the contributions for each quantity per 1°.

ΔC_X∙S = 0.0044∙(-12) = – 0.0528

ΔC_ZF∙S = – 0.04564∙(-12) = 0.54768

ΔC_ZR∙S = 0.00068∙(-12) =- 0.00816

The following procedure is just to add those quantities on C_X∙S, C_ZF∙S and C_ZR∙S of the baseline conditions, which summarized in Table 5.

	CX∙S	CZF∙S	CZR∙S	CZT∙S	Fbal	α	β
Base	1.330	– 2.280	-2.390	– 4.670	48.8	–	–
FW 15°	1.263	-1.612	-2.400	– 4.012	40.2	– 15	–
FW 12°	1.277	– 1.732	-2.398	– 4.130	41.9	– 12	–

Finally, for the first question, it is required a 12° FW angle to reach 42% target F_bal.

Question 2 – Re-balance and re-drag with α and β

Given the data on Figure 1 and 2, define which are the front and rear wing angles α and β to provide a target F_bal and C_X∙S of 42% and 1.3 m², respectively.

Figure 2 summarizes the data of RW. Since the slopes are already known, and the slope ΔF_bal/Δβ = – 0.3 %/°. This is an important information, because ΔF_bal/Δα ≃ – 0.6 %/°, thus it is possible verify that α and β should vary in ratio 1:2 to keep the car balanced.

	CX∙S	CZF∙S	CZR∙S	CZT∙S	Fbal	α	β
Base	1.277	– 1.732	-2.398	– 4.130	41.9	– 12	–

According to Table 6, the baseline configuration now is the last one with α = – 12°. Since F_bal is already 42%, the objective is to bring C_X∙S to 1.3. Adding 1° on the front wing (Table 3), the base conditions go to:

	CX∙S	CZF∙S	CZR∙S	CZT∙S	Fbal	α	β
Base	1.277	– 1.732	-2.398	– 4.130	41.9	– 12	–
FW 11°	1.281	– 1.778	– 2.397	– 4.175	42.5	– 11	–

As can be seen, C_X∙S is near to 1.3 and F_bal is 0.6 above the target. Now it is interesting to use the rear wing polar. Since FW varied in 1°, RW should vary in 2°.

	CX∙S	CZF∙S	CZR∙S	CZT∙S	Fbal	α	β
Base	1.277	– 1.732	-2.398	– 4.130	41.9	– 12	–
FW 11°	1.281	– 1.778	– 2.397	– 4.175	42.5	– 11	–
Rebal	1.301	– 1.762	– 2.443	– 4.205	41.9	-11	2

Finally, both targets are reached and the car was kept balanced.

Question 3 – Evaluate if the result of the question 2 represents an improvement

For this question it is necessary to evaluate the efficiency, which is a parameter that defines how good is the setup in providing downforce at a low cost in drag. The baseline efficiency is give by:

Eff_baseline = C_ZT∙S/C_X∙S = – 4.670 / 1.330 = – 3.51128 = – 3.5

The parameter Eff* is known as the re-balanced marginal efficiency, it can be calculated by the following equation:

Eff* = ΔC_ZT∙S/ΔC_X∙S

ΔC_ZT∙S = [- 4.205 – (- 4.670)] = 0.465

ΔC_X∙S = (1.301 – 1.330) = – 0.029

Eff* = 0.465/-0.029 = – 16.03448 ≃ – 16

Figure 3 illustrates the line between these configurations, the slope of this line is given by Eff* = – 16. The objective is the evaluation of the new configuration, if this represents an improvement. Eff* is the parameter that is compared with Eff*, which is a predetermined project constraint.

Figure 4 illustrates what Eff* represents graphically, the slope between the new option and the baseline configuration. However, Eff* should be compared to the predefined project constraint Eff*, which is parameter calculated by specific programs for simulating lap times. This estimate a re-balanced marginal efficiency for the same lap time in a given race track.

An aerodynamic option is defined better than the baseline if is characterized by a marginal efficiency that its ΔC_XS and ΔC_ZS put the new configuration inside the blue region seen in Figure 5. The problem is that neither all situations there is a lap time simulator as a support. Hence, it is possible to analyze again the Figure 3. According to Figure 5 it is possible notice that if ΔC_XS < 0, the new configuration goes to the left side side of the graph, while ΔC_ZS < 0, the new configuration goes upwards. The results for the calculations of this exercise presented and ΔC_XS = – 0.029 and. ΔC_ZS = 0.465.

As can be seen, the new configuration may not represents an improvement, it would be around the region where the wine red point indicates, thus outside the region of negative ΔC_XS and ΔC_ZS.