# Conversion from logical to sym is not possible

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Ryan 2023 年 5 月 2 日

clc; clear;
k = 10000 ; % N/m
m = 331 ; % kg
v_3 = 45 * 1609.34 * (1/3600) ; % mph to m/s
t_reach = 25/v_3 ; % s
% Using Hooke's law because the distance and stiffness are known, F = kx
F_step = k*0.05 ;
syms t
F_impulse = F_step*heaviside(t-t_reach) ;
% Make plot
% Say plot for 10 seconds like 3.2
figure(21)
fplot(F_impulse,[0,10])
title('Force Profile')
xlabel('Time (s)')
ylabel('Force (N)')
xlim([0,10])
ylim([0,550])
c_3 = 300 ;
m = 331 ;
k = 10000 ;
t_reach = t_reach ;
v_3_simple = 45 .* 1609.34 .* (1./3600) ;
% define the time interval and the number of integration steps
n = 100000;
t_3 = zeros(n,1);
tf = 10;
t_3(1) = 0;
delt = (tf-t_3(1))/n;
% initializing variables
x = zeros(n,1);
v = zeros(n,1);
x(1) = 0 ;
v(1) = v_3_simple ;
% start the integration directly from the second step (with i=2)
for i=2:1:n
% update the velocity and acceleration for x at step i
if t(i-1) >= t_reach
F = 500;
else
F = 0 ;
end
dx_3 = v(i-1) ;
dv_3 = -(c_3./m).*dx_3 - ((k./m).*x) ;
% integrate x and v
x(i) = x(i-1) + dx_3.*delt ;
v(i) = v(i-1) + dv_3.*delt ;
% integrate time
t_3(i) = t_3(i-1) + delt ;
end % end of each integration step
Conversion to logical from sym is not possible.

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### 回答 (1 件)

Walter Roberson 2023 年 5 月 3 日

syms t
That creates t as a 1 x 1 variable at the MATLAB level that contains information that links the MATLAB-level t to a variable named t that lives inside the Symbolic Engine of the Symbolic Toolbox
for i=2:1:n
% update the velocity and acceleration for x at step i
if t(i-1) >= t_reach
At that point, t at the MATLAB level is still the 1 x 1 scalar reference to the symbolic t . i is 2, so i-1 is 1, and so you are indexing the 1 x 1 symbolic t at index 1, getting back a 1 x 1 reference to the scalar symbolic t variable in the symbolic engine.
So t(i-1) at the MATLAB level resolves to the symbolic variable t that lives in the Symbolic Engine. Comparing that symbolic variable to t_reach gives a symbolic expression as a result, since symbolic engine t has no particular value, so you cannot say whether t > t_reach
Now you have if trying to figure out whether t >= t_reach is true or not. But that is not something that can be decided since t does not have a particular value. So the if fails with the error message you get.
Note that you never assign to t(i) so on the second iteration when t = 3, t(i-1) is not going to exist.

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