Combine multiple if statements for something more compact

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DIP on 19 Feb 2018
Commented: Roger Stafford on 22 Feb 2018
Friends,
I'm trying to refine my code.It works fine but I have four if conditions which I want to make more efficient. Is there an alternative way to do it?
i=1;
while (VMPH<=60)
% Vehicle speed
t(i+1) = t(i)+delt;
Vmps(i+1) = Vmps(i)+((delt*(Facc(i)))/Vm);
VMPH(i+1) = Vmps(i+1)/0.44704;
% Vehicle forces
Fr(i+1) = Fr(1);
Fd(i+1) = 0.5*Af*Cd*(Vmps(i+1))^2;
% Speed conditions
ig(i+1) = 3.78;
N(i+1) = Vmps(i+1)*io*ig(i+1)*60/(pi*Dt);
if N(i+1) > 2150
ig(i+1) = 2.06;
N(i+1) = Vmps(i+1)*io*ig(i+1)*60/(pi*Dt);
end
if N(i+1) > 2150
ig(i+1) = 1.58;
N(i+1) = Vmps(i+1)*io*ig(i+1)*60/(pi*Dt);
end
if N(i+1) > 2150
ig(i+1) = 1.21;
N(i+1) = Vmps(i+1)*io*ig(i+1)*60/(pi*Dt);
end
if N(i+1) > 2150
ig(i+1) = 0.82;
N(i+1) = Vmps(i+1)*io*ig(i+1)*60/(pi*Dt);
end
% Power and performance
Tao_b(i+1) = interp1(Speed,Torque,N(i+1));
Ft(i+1) = Tao_w(i+1)/Dt*2;
Pb(i+1) = 2*pi*Tao_b(i+1)*N(i+1)/60;
% Acceleration force
Facc(i+1) = Ft(i+1)-Fd(i+1)-Fr(i+1);
i=i+1;
end
Thank You!
M on 21 Feb 2018
Why not ?

Jeff Miller on 19 Feb 2018
Maybe something like this:
IGVals = [3.78 2.06 1.58 1.21 0.82];
CurrentIG = 1;
i=1;
while (VMPH<=60)
...
% Speed conditions
ig(i+1) = IGVals(CurrentIG);
N(i+1) = Vmps(i+1)*io*ig(i+1)*60/(pi*Dt);
if N(i+1) > 2150
CurrentIG = CurrentIG + 1;
ig(i+1) = IGVals(CurrentIG);
N(i+1) = Vmps(i+1)*io*ig(i+1)*60/(pi*Dt);
end
% Power and performance
...
DIP on 21 Feb 2018
Edited: DIP on 21 Feb 2018
Jeff, wouldnt the line
ig(i+1) = IGVals(CurrentIG);
throw dimension mismatch error ? Edit : It does throw a index exceeds bounds error.

Roger Stafford on 19 Feb 2018
Edited: Roger Stafford on 19 Feb 2018
You can replace the part of the code after "%Speed Conditions" but before "% Power and performance" by these lines:
t = Vmps(i+1)*io*60/(pi*Dt);
x = [3.78,2.06,1.58,1.21,0.82];
ig(i+1) = x(sum(2150<(x(1:4)*t))+1);
N(i+1) = ig(i+1)*t;
They should produce an equivalent result.
Roger Stafford on 22 Feb 2018
@DIP: That line should work. The expression
sum(2150<(x(1:4)*t))+1
should provide an integer value ranging from 1 to 5. This in turn should be a valid index in the vector x. You can do some checking by writing
ix = sum(2150<(x(1:4)*t))+1;
disp(ix)
to display the values of ix. By the way, it is assumed that the variable t is a scalar. If not, you would probably get a similar error message at the ix calculation.

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