Problem 3064. Cycling — Normalized Power
Solution Stats
Problem Comments
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1 Comment
I don't understand what I'm looking at, Ive attempted to add the code from a solution, but it seems like I'm missing something crutial...
function [P_avg,NP] = cycling_norm_power(power)
P_avg = 0;
NP = 0;
%% steady
power = 200*ones(1,3600);
P_avg_corr = 200;
NP_corr = 200;
[P_avg,NP] = cycling_norm_power(power);
assert(isequal(P_avg_corr,P_avg))
assert(isequal(NP_corr,NP))
Solution Comments
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2 Comments
??????????? really? 4 correct and 5 not?
Same problem as noted in the other comment. You get the correct P_avg here but not NP because the sprint is so short that the 30-second rolling average (applied twice) is rounded to the same answer. Unrounded P_avg answers for this routine and the correct routine are 131.3638 and 131.1111, respectively. Given that answers are being rounded, it may look wrong since the P_avg calculation appears to be right (due to rounding) while NP was wrong when both were actually wrong.
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2 Comments
Maybe I'm wrong, but for sure the results in test 4 and 5 is NOT CORRECT
Note in the last paragraph of the explanation: "You will be provided with the 30-second rolling average power data set (vector)." In essence, your solution is calculating the 30-second rolling average of a 30-second rolling average data set.
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