Why matlab make this calculation error?
显示 更早的评论
for this code i wrote for my home work, when i was running, SOME of the output from line 17 is not correct. Can any one explain to me why this is not correct? (I use number: 38 3,-8.9,4.1 as 4 sample input)
T = input('please enter the force: '); %%1
x = input('please enter the x coordinate: '); %%2
y = input('please enter the y coordinate: '); %%3
z = input('please enter the z coordinate: '); %%4
fprintf('\n');%%5
fprintf('\n'); %%6
fprintf(' x = %4.2f , y = %4.2f, z= %4.2f', x,y,z);%%7
fprintf('\n'); %%7
fprintf('\n'); %%8
R = sqrt(x^2+y^2+z^2); %%9
norm_x = x/R; %%10
norm_y = y/R;%%11
norm_z = z/R;%%12
fprintf('||dist|| = %4.2f',R); %%13
fprintf('\n'); %%14
fprintf(' norm= %4.3f i+%4.3fj+ %4.3f k', norm_x,norm_y,norm_z); %%15
fprintf('\n'); %%16
fprintf('T = %4.3f i + %4.3f j + %4.3f k',T*norm_x, T * norm_y, norm_z*T);%%17%%The first two output has errors.
fprintf('\n');%%18
The incorrect output is shown here, it should be 11.134, -32.984, 15.2
T = 11.124 i + -33.002 j + 15.203 k
2 个评论
Star Strider
2016-2-24
Are you sure your ‘correct’ values are in fact correct?
I see no obvious errors, and if the norm values in Line 15 are correct, and ‘T’ is correctly entered, Line 17 should be correct.
BEICHEN LIU
2016-2-24
回答（1 个）
Walter Roberson
2016-2-24
0 个投票
No, the output is correct and it is your expected output that is wrong. In particular, when you calculated the norm components, you rounded to three decimal places before doing the multiplication by the force, and that gave you your "expected" answer; when you do not round to three decimal places you get the actual answer the code shows.
4 个评论
BEICHEN LIU
2016-2-24
@BEICHEN LIU: MATLAB is not making any mistakes, although your title accuses it of this. In fact you are doing two different calculations which, not surprisingly, give two different results:
>> T*0.293
ans = 11.134
>> T*norm_x
ans = 11.124
You are confusing the rounded printed value of norm_x for the actual stored numeric value of norm_x. These are, however, clearly not the same, no matter how much you imagine them to be.
Roger Stafford
2016-2-24
@Beichen Liu: Please read Walter's explanation more carefully! Multiplying 38 by .293 is incorrect, because the actual value of norm_x is .292742143, and when you multiply that by 38, you get 11.12420143 which rounds to 11.124, not 11.134. You were misled by your own intermediate rounding displays using 'fprintf'.
Steven Lord
2016-2-24
You can easily check to convince yourself that Walter and Stephen are correct. Display (using DISP, not using FPRINTF) the value 0.293-norm_x. That value will be nonzero. [It's probably going to be somewhere in the vicinity of 2.6e-4 based on T having a value of 38.]
类别
在 帮助中心 和 File Exchange 中查找有关 MATLAB 的更多信息
产品
2016-2-24
2016-2-24
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
