In this post, we analyse the additional pieces of information in turn and introduce labels for the main quantities and key points in time.
1) We want to stop at exactly 10 miles so we are ready to turn around to go backThis is the main objective or goal of the problem. To precisely formulate the problem, let's express this objective in symbols. We also need to convert to metric units (e.g. metres) rather than imperial (miles), and to decide where to measure from. The most obvious choice is to measure from the start of the 10 mile track, although as this problem is focussed on the stopping phase an alternative would be to measure distance from the end of the measured mile. For now, we'll let s be the distance (in metres, m) from the start line, and v be the velocity (in metres per second m/s) at time t (in seconds). We can use an online calculator to work out that 10 miles is 16093 metres (or 16.093 km). We can then write this objective in symbols as
v = 0 m/s, when s = 16093 m
2) We can not deploy the parachutes above a certain speed, definitely not supersonicThis is a constraint on the timing of when the parachutes can be deployed. We'll label these points in time so we can refer to them more easily. Let t1 be the time in seconds when the first parachute is deployed and t2 be the time in seconds when the second parachute is deployed. The velocity at t1 is written v(t1) and must be less than supersonic (i.e. less than the speed of sound). The speed of sound depends on the environmental conditions, but is approximately 343 m/s. This constraint can be written in symbols as
v( t1 ) < 'a certain speed' < 343
3) We can stop at 10 miles by two methods
a) Airbrake at the end of the measured mile and wheel brakes at about 150 mph. This is the preferred option.
b) Parachutes: first just sub-sonic, second almost immediately afterwards, followed by wheel brakes maybe a bit faster than 150 mphThis piece of additional information suggests that we should split the problem into 2 cases, with and without the air-brake.
Let t0 be the time when the air-brake is deployed, and t3 be the time when the wheel brakes are applied. We have chosen the time labels to reflect the expectation that t0 <= t1 <= t2 <= t3. Let s0 be the distance (in metres) of the end of the measured mile from the start line.
In case (a), BLOODHOUND is stopped using the air-brake at the end of the measured mile, followed by the wheel brakes when the velocity is approximately 150 mph (or 67.06 m/s ). In symbols these constraints can be written as
s(t0) = s0 and v(t3) is approximately 67.06 m/s
In case (b), the constraints can be written as
v(t1) = 343 - (a bit), v(t2) = 343 - (a bit more), v(t3) = 67.06 + (another bit)
where 'a bit','a bit more' and 'another bit' are small positive velocities that need to be decided.
4) If the airbrake only partly deploys we will need one parachute, but not necessarily at high speed. This will be very dynamicThis sounds like an interesting case. Let's call it case (c). In this case we need to find the velocity when the first parachute is deployed, that is v(t1).
5) The air-brake has one good mode of deployment and four possible failure modes, three with a reduced function, one with no functionThis point introduces more logic for us to handle. Based on the language John uses it sounds like the good mode of deployment will put us into case (a), the 'no function' mode puts us into case (b), and the 3 reduced function failure modes all put us into case (c). Let's call the 3 reduced function cases (c1), (c2) and (c3).
Analysing these additional pieces of information has highlighted the importance of understanding the velocity, v(t), and the distance s(t). In the next blog, we'll start building up a model for the velocity of BLOODHOUND.
