Need help on Matlab grader problem skydiver

The code:

[height,time]=skydiverdatagen

will generate test data of a skydiver falling. The values of height will be measured in meters and they will be positive except for the last one which is 0. The values for time are evenlyspaced and start at 0 and the rest are positive.

From this data you need to calculate several quantities

1 Terminal Velocity before the parachute is pulled. This is simply the largest velocity during the entire the jump. Take the absolute value so that you get a positive value. You must store this as:

TV1

2 Terminal Velocity after the parachute is pulled. This is velocity that the skydiver will hit the ground at. In other words, this is the final velocity.Take the absolute value so that you get a positive value. You must store this as:

TV2

3 The time the parachute was pulled. That is the moment in time when acceleration first becomes positive. You must store this as:

PT

4 The maximum acceleration experienced by the skydiver prior to hitting the ground. Take the absolute value so that you get a positive value. You must store this as:

MA

Each of these values, must be calculated using central difference derivatives. You answers will need to be accurate for full credit.

I can get the TV1 TV2 and PT but cannot get the MA correct .

I currently have this code:

[height,time]=skydiverdatagen;
velocity = zeros(size(height));
velocity(2:end-1) = (height(3:end) - height(1:end-2)) ./ (time(3:end) - time(1:end-2));
acceleration = zeros(size(height));
acceleration(2:end-1) = (velocity(3:end) - velocity(1:end-2)) ./ (time(3:end) - time(1:end-2));
TV1 = max(abs(velocity));
TV2 = abs(velocity(end-1));
PT = time(find(acceleration > 0, 1, 'first'));
MA = max(abs(acceleration(1:end-1))); 

[Walter Roberson]

velocity(2:end-1) = (height(3:end) - height(1:end-2)) ./ (time(3:end) - time(1:end-2));

This is not the Foreward Difference, or the Backwards Difference, or the Central Difference. It is closest to the Central Difference, but the central difference requires dividing the above result by 2.

I do not think you have taken correct numeric gradients.

Have you considered calling gradient ?

[Logan]

We have not been intoduced to gradient in my class but I just don't really understand how to use central difference here and what the "h" value would be in this scenario

[Logan]

would the h just be 1 and this is what you mean:

[height,time]=skydiverdatagen;
velocity = zeros(size(height));
velocity(2:end-1) = ((height(3:end) - height(1:end-2))/2) ./ ((time(3:end) - time(1:end-2))/2);
acceleration = zeros(size(height));
acceleration(2:end-1) = ((velocity(3:end) - velocity(1:end-2))/2) ./ ((time(3:end) - time(1:end-2))/2);
TV1 = max(abs(velocity));
TV2 = abs(velocity(end-1));
PT = time(find(acceleration > 0, 1));
PT_index = find(acceleration > 0, 1);
MA = max(abs(acceleration(1:PT_index-1)))

[Walter Roberson]

https://www.sheffield.ac.uk/media/32080/download?attachment discusses the Central Difference Formula

[Logan]

My homework is still saying my MA is wrong but every other vairable is calculated correctly I do not understand what the issue is?

[height,time]=skydiverdatagen;
velocity = zeros(size(height));
velocity(2:end-1) = (height(3:end) - height(1:end-2)) ./ (2 * (time(2) - time(1)));
acceleration = zeros(size(height));
acceleration(2:end-1) = (velocity(3:end) - velocity(1:end-2)) ./ (2 * (time(2) - time(1)));
TV1 = max(abs(velocity(1:find(acceleration > 0, 1) - 1)));
TV2 = abs(velocity(end-1));
PT_index = find(acceleration > 0, 1);
PT = time(PT_index);
MA = max(abs(acceleration(1:PT_index-1)));

[Vidhi Agarwal]

Hi @Logan,

I understand that the maximum acceleration, is likely causing the issue. The code for calculation of "MA" is:

MA = max(abs(acceleration(1:PT_index-1)));

• The acceleration computed at index i uses velocity(i+1) and velocity(i-1), so the value at acceleration(i) corresponds to the time at time(i).
• When you do acceleration(1:PT_index-1), you are excluding the acceleration at the moment the parachute deploys, which is probably the maximum (since the acceleration spike happens when the chute opens).
• The maximum acceleration experienced is typically at the moment the parachute opens.
• So try including up to and including PT_index.

So the change code for "MA" calculation is given below:

MA = max(abs(acceleration(1:PT_index)));

Hope this helps!