Find a variable from an integration equation
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Hi,
I want to calculate motor_power variable from know total_time variable.
All other parameters are known expect motor_power.
I have tried fsolve and fzero functions but throwing errors related to integration
Kindly request you to support for my problem
The code is as below:
ct=@(speed)base_speed1./(motor_power-(a+b.*speed.^2).*base_speed); %Function 1%
cp=@(speed)speed./(motor_power-(a+b.*speed.^2).*speed); %Function 2%
total_time=m_veh*integral(ct,0,base_speed)+m_veh*integral(cp,base_speed,max_speed); %Integration equation%
3 个评论
Matt J
2023-9-1
Your posted code does not contain fsolve or fzero or the error messages they threw.
Dyuman Joshi
2023-9-2
"I have tried fsolve and fzero functions but throwing errors related to integration"
You have to show what you tried with those functions and what errors did you get. Copy and paste the rest of the code and the error messages you get i.e. all of the red text.
Harsh
2023-9-2
采纳的回答
Torsten
2023-9-1
ct=@(speed,motor_power)base_speed1./(motor_power-(a+b.*speed.^2).*base_speed); %Function 1%
cp=@(speed,motor_power)speed./(motor_power-(a+b.*speed.^2).*speed); %Function 2%
fun = @(motor_power)total_time-(m_veh*integral(@(speed)ct(speed,motor_power),0,base_speed)+m_veh*integral(@(speed)cp(speed,motorpower),base_speed,max_speed));
motor_power0 = 10; % initial guess
motor_power = fsolve(fun,motor_power0)
9 个评论
Harsh
2023-9-2
clear all
clc
I thoguht motor_power is unknown and the problem is to find its value?
% motor_power=239000; to be solved?
r=0.344;
s=1;
original_base_speed=20.3;
b=original_base_speed/s;
m=27.77;
m_v=1700*1.12;
g=9.81;
m_1=0.006;
m_2=0.0001;
r=1.204;
c_d=0.23;
height=1.59;
width=1.627;
area= 2.22;
total_time=5.68;
if b<=m
b_speed=b;
else
b_speed=m;
end
ct=@(speed)b./(motor_power-(m_v*g.*(m_1+m_2.*speed)+0.5*r*c_d*area.*speed.^2).*b_speed);
if m>b
cp=@(speed)speed./(motor_power-(m_v*g.*(m_1+m_2.*speed)+0.5*r*c_d*area.*speed.^2).*speed);
else
cp=0;
end
ct
cp
ct and cp are both functions of one variable, speed. However, in the defintion of fun they are both called with two input arguments, speed and motor_power.
Perhaps ct and cp should be defined with motor_power as a second argument?
fun = @(motor_power)total_time-(m_v*integral(@(speed)ct(speed,motor_power),0,b_speed)+m_v*integral(@(speed)cp(speed,motor_power),b_speed,m))
r=0.344;
s=1;
original_base_speed=20.3;
b=original_base_speed/s;
m=27.77;
m_v=1700*1.12;
g=9.81;
m_1=0.006;
m_2=0.0001;
r=1.204;
c_d=0.23;
height=1.59;
width=1.627;
area= 2.22;
total_time=5.68;
if b<=m
b_speed=b;
else
b_speed=m;
end
ct=@(speed,motor_power)b./(motor_power-(m_v*g.*(m_1+m_2.*speed)+0.5*r*c_d*area.*speed.^2).*b_speed);
if m>b
cp=@(speed,motor_power)speed./(motor_power-(m_v*g.*(m_1+m_2.*speed)+0.5*r*c_d*area.*speed.^2).*speed);
else
cp=@(speed,motor_power)zeros(size(speed));
end
fun = @(motor_power)total_time-(m_v*integral(@(speed)ct(speed,motor_power),0,b_speed)+m_v*integral(@(speed)cp(speed,motor_power),b_speed,m));
motor_power0 = 2e5; % initial guess
motor_power = fsolve(fun,motor_power0,optimset('MaxFunEvals',10000,'MaxIter',10000))
% Plot control
motor_power = 100000:100:400000;
plot(motor_power,arrayfun(@(motor_power)fun(motor_power),motor_power))
Harsh
2023-9-2
the equation behind it is as below where p is the motor power and R is the resistance which depends on the velocity V 
Torsten
2023-9-2
It's solved. Didn't you see the output from "fsolve" : 202 540 ?
Harsh
2023-9-2
Q = @(v) sym(v);
r = Q(344)/Q(1000);
s = Q(1);
original_base_speed = Q(203)/Q(10);
b = original_base_speed/s;
m = Q(2777)/Q(100);
m_v = Q(1700)*Q(112)/Q(100);
g = Q(981)/Q(100);
m_1 = Q(0.006);
m_2 = Q(0.0001);
r = Q(1204)/Q(1000);
c_d = Q(23)/Q(100);
height = Q(159)/Q(100);
width = Q(1627)/Q(1000);
area = Q(222)/Q(100);
total_time = Q(568)/Q(100);
if b<=m
b_speed=b;
else
b_speed=m;
end
syms speed motor_power
ct(speed,motor_power) = b./(motor_power-(m_v*g.*(m_1+m_2.*speed)+0.5*r*c_d*area.*speed.^2).*b_speed);
if m>b
cp(speed,motor_power) = speed./(motor_power-(m_v*g.*(m_1+m_2.*speed)+0.5*r*c_d*area.*speed.^2).*speed);
else
cp(speed,motor_power) = zeros(size(speed));
end
fun(motor_power) = total_time-(m_v*int(ct(speed,motor_power),speed,0,b_speed)+m_v*int(cp(speed,motor_power),speed,b_speed,m))
limit(fun(motor_power), motor_power, -inf)
df = diff(fun(motor_power));
limit(df, motor_power, -inf)
The lower limit is positive, and the derivative is positive out to negative infinity... so the function has no zero.
Possible exception would be in the range that is covered by the piecewise, but the function is everywhere positive in that range too.
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