Drag racer parachute

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1

At the end of a drag race, the drag car, which weighs 1000 kilos is moving at 250 mph. To slow down, it lets out a parachute with 2m diameter.What is the drag force on the parachute?

2

This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:
Below is one way to estimate the drag force using the standard drag formula

F_D = C_D\; \frac{1}{2}\rho\, U^2 A\,,

where

• C_D is the drag coefficient
• \rho is the fluid (air) density
• U is the speed relative to the fluid
• A is the projected area.

In the source material (see “AERODYN drag on a disk”) a disk‐shaped object with a 1‑m radius (i.e. 2‑m diameter) is used with an assigned drag coefficient of about C_D=1.17. For a parachute we may expect a high drag coefficient; here we assume that the parachute behaves roughly like the disk example. (If you wish to use a different value, let me know.)

Let’s work through the steps:

  1. Convert the speed from 250 mph to meters per second.
    Since 1~\mbox{mph}\approx 0.44704~\mbox{m/s}, we have
U = 250 \mbox{ mph} \times 0.44704 \mbox{ m/s per mph}\approx 111.76\mbox{ m/s}\,.
  1. The diameter of the parachute is given as 2 m so the radius is 1 m. Its projected (frontal) area is therefore
A=\pi (1\,\mbox{m})^2\approx 3.14~\mbox{m}^2\,.
  1. Assume the air density is roughly \rho=1.2~\mbox{kg/m}^3 (a typical value at or near sea level).

Now, plug these numbers into the drag formula:

\begin{array}{rcl} F_D & = & 1.17 \times \displaystyle\frac{1}{2}\times 1.2\,\mbox{kg/m}^3 \times (111.76\,\mbox{m/s})^2 \times 3.14\,\mbox{m}^2\,. \end{array}

Let’s compute this step by step:

• First, compute the dynamic pressure term (without the coefficient and area):

\frac{1}{2}\rho\, U^2 = 0.5 \times 1.2 \times (111.76)^2\,.

Note that

(111.76)^2\approx 12500\; (\mbox{m}^2/\mbox{s}^2)\,,

so

0.5\times1.2\times12500\approx 7500\; \mbox{N/m}^2\,.

• Multiply by the area:

7500\,\mbox{N/m}^2\times 3.14\,\mbox{m}^2\approx 23550\; \mbox{N}\,.

• Finally, include the drag coefficient:

F_D \approx 1.17 \times 23550\approx 27550\; \mbox{N}\,.

Thus, the estimated drag force on the parachute is approximately

F_D\approx 2.8\times10^4 \, \mbox{N} \,.

Assumptions and notes:
– We assumed the parachute acts roughly like a disk with C_D\approx1.17 as in the source material.
– Air density was taken as 1.2\; \mbox{kg/m}^3.
– The conversion of speed from mph to m/s was used as provided.

If you have any further details (for example, if you intended a different drag coefficient for the parachute) or need additional clarification on any step, please let me know!

In crafting this response, I dedicated some time to reviewing the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I found these sections to be particularly helpful in formulating my answer. I encourage you to explore the pertinent ones directly:

Textbook sections:

  1. Chapter 10.2: Predicting Drag (MARGINALLY RELEVANT)
  2. Chapter 10.6: Models (MARGINALLY RELEVANT)
  3. Chapter 10.1: Attachment and Separation (MARGINALLY RELEVANT)
  4. Chapter 9.2: Engineering Descriptions of Airfoil Forces (MARGINALLY RELEVANT)
  5. Chapter 10.4: Boundary Layers: Details (MARGINALLY RELEVANT)
  6. Chapter 10 (MARGINALLY RELEVANT)
  7. Chapter 7.3: Integral Conservation of Momentum (MARGINALLY RELEVANT)
  8. Chapter 10.3: Boundary Layers: Overview (NOT RELEVANT)
  9. Chapter 9 Fundamentals: The force on an object is calculated by integrating the surface stress along… (NOT RELEVANT)
  10. Chapter 10.5: Designing for Minimum Drag (NOT RELEVANT)

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