Air is not just “in the way”. It is a force you can work out from density, speed, shape and area. This pack uses the same drag equation we check our SimScale results against, and it settles an argument we have had all season: is it better to cut frontal area or to cut Cd?
Explain why doubling the speed makes the drag four times bigger, and sketch the drag against speed curve.
03
Link Cd and frontal area to the shape decisions on a STEM Racing car.
04
Show with algebra that cutting A by 10% and cutting Cd by 10% reduce drag by the same amount.
02From the Nightmare car
Our aero work is not a poster. We change the body, run it in SimScale, and ask one question: did CdA go down? The Aero Overlay Card we take to school stands shows where the air speeds up and where pressure builds on a side view of the car. The QR code on the back opens this pack so a student can do the maths behind the picture. Every aero review this season ended with this equation.
Fig. 3 Drag comes from the pressure difference between the nose and the wake, plus friction along the body. Shape (Cd) and size (A) are the things we can design. The v² term is why top speed costs so much.
03Core idea
Fd = ½ · ρ · v² · Cd · A
ρ is the density of the air in kg/m³. Use 1.20 kg/m³ unless you are given another value.
v is the speed through the air in m/s. It is squared, so twice the speed gives four times this part of the product.
Cd is the drag coefficient. It has no unit. Lower means a slipperier shape. Our SimScale runs report this number.
A is the frontal area in m². To change cm² into m², divide by 10 000.
Why v² hurts so much
Two things grow with speed: how many air molecules you hit each second, and how hard you hit each one. Both are proportional to v, so drag is proportional to v × v. It also means the last few metres of a run, at top speed, are where the most energy is lost.
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Engineers compare the product CdA. Both terms sit in the equation the same way, so a 10% cut in either one gives the same 10% cut in drag. Q3 asks you to prove it, and the simulator’s compare curve shows it.
04Drag calculator and the v² curve
Set the four inputs and read off Fd. The chart draws drag against speed for your car and for a compare car. Choose whether the compare car cuts area or Cd by 10%, and watch the two curves land on top of each other.
The values are realistic for a STEM Racing car but they are not a CFD result. If you want the real numbers, use the SimScale sheet.
—½ρv² (Pa)
—CdA (m²)
—NFd, the product
—WPower = Fd × v
05Worked example
Given: ρ = 1.20 kg/m³, v = 12 m/s, Cd = 0.28, A = 0.0045 m².
½ρ = 0.60
v² = 144
½ρv² = 0.60 × 144 = 86.4
CdA = 0.28 × 0.0045 = 0.00126
Fd = 86.4 × 0.00126 = 0.109 N (3 s.f.)
If a body change takes Cd from 0.28 down to 0.25 at the same area and speed, Fd is multiplied by 0.25/0.28, which is about 0.89. That is an 11% cut in drag before you touch the speed. At 20 m/s the same car feels 0.302 N and is spending about 6 W pushing air out of the way. That is why the last five metres decide the race.
06Student questions Teacher mode is off. Answers are hidden.
Q1 · Substitute[3]
ρ = 1.20, v = 10 m/s, Cd = 0.30, A = 0.0050 m². Work out Fd. Show each product on its own line.
Answer key
½ρv² = 0.5 × 1.20 × 100 = 60
CdA = 0.30 × 0.0050 = 0.0015
Fd = 60 × 0.0015 = 0.090 N
Q2 · Reason[2]
Keep ρ, Cd and A the same. Compare Fd at 8 m/s and at 16 m/s. What factor links them, and why?
Answer key
A factor of 4 (1). The speed doubles and drag depends on v², so (16/8)² = 4 (1). Accept “four times” with the reasoning.
Q3 · Design[3]
You can either cut A by 10% or cut Cd by 10%. Which change reduces Fd more? Prove it with algebra.
Answer key
Neither. They are the same. Write Fd = k·Cd·A where k = ½ρv² (1). Cutting A gives k·Cd·(0.9A) = 0.9Fd. Cutting Cd gives k·(0.9Cd)·A = 0.9Fd (1). Both terms are linear, so a 10% cut in either one gives a 10% cut in drag. The design choice comes down to which one is easier to build (1).
Q4 · Extension[3]
At top speed, v = 20 m/s, CdA = 0.00126 m² and ρ = 1.20. Find Fd and the power being spent against the air (P = Fd × v).
Answer key
½ρv² = 0.6 × 400 = 240
Fd = 240 × 0.00126 = 0.302 N
P = 0.302 × 20 = 6.05 W
07Printable worksheet The answer key prints as well when teacher mode is on
Team Nightmare · STEM Subject Lab
NM-LAB-03 · Drag Maths
Name ______________________ Class __________ Date __________
Accept any equivalent working. The marks in brackets match the student sheet.
Teacher copy Do not hand out
1. [4] ½ρv² = 0.5 × 1.20 × 121 = 72.6CdA = 0.27 × 0.0048 = 0.001296Fd = 72.6 × 0.001296 = 0.0941 N (3 s.f.) One mark for each product line and one for the final answer with a unit.
2. [2] Drag rises with v², so it is biggest exactly when the car is fastest (1). The force acts over the whole lane and most of the energy is lost near top speed, which decides the finish time (1).
3. [2] A curve through the origin that gets steeper (1). Fd at 20 m/s is four times Fd at 10 m/s (1).
4. [3] ½ρv² = 86.4 (from the worked example)Fd = 86.4 × 0.00110 = 0.0950 NDrop = (0.1089 − 0.0950)/0.1089 = 12.7%, about 13%