On Twitter, @printerelf, an aerospace engineer, suggested a couple of good questions. He asked

Q1. How much KE (Kinetic Energy) will the car have when the chutes are deployed?

Q2. How big are the parachutes?

Q2. How big are the parachutes?

Let's take these questions in turn. Firstly, the definition of kinetic energy is 0.5*mass*velocity

^{2}, therefore, in order to know the kinetic energy of BLOODHOUND, we need to know its mass and velocity. We can look up that the mass of the car plus the driver Andy Green (once the fuel has burnt off) will be approximately 5000 kg, with more details available from http://www.bloodhoundssc.com/car/facts_and_figures.cfm. However, the velocity of the car when the chutes are deployed is unknown, in fact, the velocities can be thought of as design parameters that we can adjust to find the best way of stopping BLOODHOUND. Hence, after a bit of thought, we see that Q1 is just an alternative way of phrasing the part of the initial problem (i.e. where to place the markers around the Speedo to give Andy guidance of when to deploy the parachutes).Re-phrasing vaguely worded problems can often help to understand the problem better, or to suggest possible routes to a solution. In this case, the thought of looking at the problem from an energy perspective strikes me as a promising one. It prompts me to recall that many physics-related problems can be solved through applying the laws of conservation of energy, and indeed in a recent phone call with John, he'd mentioned that an energy-based analysis may be relevant. We should keep this inkling in mind as we explore the problem further.

Q2 is also pertinent. The size of the parachutes will relate to the amount of drag (the force acting to slow down BLOODHOUND): the bigger the area of the parachute, the larger the drag. However, the parachute size isn't known as the supplier for the parachutes hasn't been fixed yet. This is another typical feature of real-world problems - the numbers used in the calculations change as the project progresses. For now, John tells me we should use a 'D/q value' (also known as a drag coefficient) of 0.8 for both parachutes. I remind myself (using a spreadsheet Ron Ayers once gave me) that it is common in aerodynamics to model drag as the product of the drag coefficient 'D/q' and q, the dynamic pressure, i.e. DragChute = 0.8*q. Here, DragChute is the drag force from the chute in Newtons (N), and q is the dynamic pressure in m

^{2}.Both @printerelf's questions revealed more information about the problem in hand. What other questions might you want to ask at this stage? Suggestions welcome in the comments below.

The first questions I actually asked John reflected my background as a mathematician and were aimed at trying to cast the problem as the solution to an optimization problem. I asked:

What is the objective of the choice of chute opening times? E.g. Is it to stop at exactly 10 miles, or for them to open at defined speeds? What constraints apply?Tune in next week for John's reply...

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