Approximating /transforming /absolute error

input data

r=3;h=8;Dv=0.1;

Dr=(Dv/3)*(1/(1/3*pi*h*2*r))

the result is 6.6315e-04 , how do i transform /aproximate/etc the result on the left to 0.00066 ? I ran a conversion to decimal using Symbolab (website) and the exact result of the conversion is 0.00066315 .

Answer 1

[Alan Stevens]

For display purposes you can use fprintf:

>> r=3;h=8;Dv=0.1;
Dr=(Dv/3)*(1/(1/3*pi*h*2*r));
fprintf('%0.2g \n',Dr)
0.00066

[Opariuc Andrei]

Thank you

[Opariuc Andrei]

i have another question related to this topic. How can i write to get 0.01 precision ? or higher

[Alan Stevens]

Do you mean like this

>> r=3;h=8;Dv=0.1;
Dr=(Dv/3)*(1/(1/3*pi*h*2*r));
fprintf('%0.6g \n',Dr)
0.000663146 

Type doc fprintf in the command window for more details on controlling the output with fprintf.

[Opariuc Andrei]

i've got a rectangular parallelepiped with the base a rectangle having the sides of the base A ≈ 15 mm and B ≈ 20 m and height H ≈ 25 m. The body is drilled vertically with a drill with a diameter D ≈ 3 mm. i'm supposed to determine the absolute errors in determining the values ​​A, B, H, D so that the body volume can be calculated with a accuracy of 0.01 mm3. this is what i'm referring to .

[Alan Stevens]

That's an entirely different question! fprintf merely alters the way the number displays, it doesn't alter the size of the values that have been used internally.

If you assume the absolute error is the same on A, B, H and D, (say, delta mm), then fzero will find the value of delta that makes the body Volume have an absolute error of 0.01mm^3. For example

delta0 = 10^-5;
delta = fzero(@dfn,delta0);
disp(delta)
fprintf('delta = %0.7f mm\n',delta)

function tol = dfn(delta)

      A = 15 + delta;   % mm
      B = 20 + delta;   % mm
      H = 25 + delta;   % mm
      D = 3 + delta;    % mm
      Vtol = 0.01;      % mm^3
      
    %      Volume with error    Nominal volume       Volume tolerance  
      tol = A*B*H-pi*D^2/4  - (15*20*25 - pi*3^2/4) - Vtol;      
end

fzero adjusts delta until the value returned from dfn is as close to zero that it can get.

tol returns the difference between the volume .

If these aren't the right equations for you, then modify to suit.

[Opariuc Andrei]

Oh master , please accept my humble thank you :)

[Alan Stevens]

You're welcome. Note that I've suggested a very simplistic approach. A more advanced approach might involve a Monte-Carlo simulation and/or allow different tolerances on each of A, B etc.