Combine multiple if statements for something more compact

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));
        Tao_w(i+1)= Tao_b(i+1)*io*ig(i+1)*etadrive;
        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!

[Moritz]

Why don't you just delete all but the last one? The rest doesn't matter anyhow as it is overwritten by the last if statement.

[DIP]

Moritz, im modelling a gear box. Hence the last if statement is the final drive gear. The gear ratios go from 3.78 , 2.06, 1.58, 1.21, 0.82.

[DIP]

Im pretty sure there is a better way to code this although im getting the intended result.

[Birdman]

Does i stands for ith gear?

[DIP]

No, i is just the while loop counter. It is a 5 speed gear box - each time I hit a particular RPM, I switch gear. Hence 4 if loops for each gear shifted.

[Basil C.]

Why do all the four IF statements have the same condition that

 N(i+1)>2150

[DIP]

Basil, Each time the engine hits 2150 rpm, the gear shifts one step higher. i am an inefficient coder, somehow this gives me the right answer. However, im looking to modify the code and make it simple.

[M]

Do you ant to execute something while

N(i+1)>2150 ?

Something like

while N(i+1)>2150 && count<4
   count=count+1;
   your code here
end

[DIP]

So i define a while loop within a while loop?

[M]

Why not ?

Answer 1

[Jeff Miller]

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]

Jeff, wouldnt the line

ig(i+1) = IGVals(CurrentIG);

throw dimension mismatch error ? Edit : It does throw a index exceeds bounds error.

Answer 2

[Roger Stafford]

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.

[DIP]

Roger, i get an error Matrix dimensions must agree for the third line of your suggestion.

[Roger Stafford]

@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.