- #1
jacobi1
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A motor torque curve is given by \(\displaystyle \tau(\omega) = -\frac{\tau_{s}}{\omega_{f}} \omega +\tau_{s}\), where
\(\displaystyle \tau_{s}\) is the stall torque and \(\displaystyle \omega_{f}\) is the free angular speed. Our objective is to
find the value of the gear ratio \(\displaystyle R \) that minimizes the time required to reach a certain angular displacement
\(\displaystyle \theta \).
We begin from the equation \(\displaystyle \tau = I \frac{d \omega}{dt} \). Setting the two torque expressions equal and rearranging, we get
\(\displaystyle \frac{d \omega}{dt} + \frac{ \tau_{s} \omega}{I \omega_{f}} =
\frac{\tau_{s}}{I}\).
The integrating factor is \(\displaystyle \mu = e^{\int \frac{\tau_{s}}{I \omega_{f}} dt} = e^{\frac{\tau_{s} t}{I \omega_{f}}}
\). Multiplying all the terms of the equation by this factor and recognizing that the left side becomes a derivative, we obtain
\(\displaystyle \left ( \omega e^{\frac{\tau_{s} t}{I \omega_{f}}} \right )' = \frac{\tau_{s}}{I}
e^{\frac{\tau_{s} t}{I \omega_{f}}} \).
Integrating both sides from 0 to t and using the fact that \(\displaystyle \omega=0\) when \(\displaystyle t=0\), we obtain after rearrangement
\(\displaystyle \omega(t)= \omega_{f} \left (1-
e^{- \frac{\tau_{s} t}{I \omega_{f}}} \right )\).
Integrating again from 0 to t and using the fact that \(\displaystyle \theta=0\) when \(\displaystyle t=0\), we obtain
\(\displaystyle \theta(t)= t \omega_{f} - \frac{I \omega_{f}^2}{\tau_{s}}
\left (1 + e^{- \frac{\tau_{s} t}{I \omega_{f}}} \right)\).
Making the substitutions \(\displaystyle \tau_{s} \mapsto
\frac{\varepsilon \tau_{s}}{R}\) (where \(\displaystyle \varepsilon\) is efficiency) and
\(\displaystyle \omega_{f} \mapsto \omega_{f} R\), the equation becomes
\(\displaystyle \theta(t)= t \omega_{f} R - \frac{I \omega_{f}^2 R^3}{\varepsilon \tau_{s}}
\left (1 + e^{- \frac{\varepsilon \tau_{s} t}{I \omega_{f} R^2}} \right )\).
Next, we set
\(\displaystyle a= \omega_{f} R\),
\(\displaystyle b= \frac{I \omega_{f}^2 R^3}{\varepsilon \tau_{s}}\), and
\(\displaystyle c= \frac{a}{b}\).
The equation becomes \(\displaystyle \theta= at -b e^{-ct}-b\). We wish to invert this to get \(\displaystyle \theta\) as a function of t.
Adding b to both sides, multiplying by \(\displaystyle e^{ct}\), and combining the \(\displaystyle e^{ct}\) terms, we have
\(\displaystyle (at- \theta -b) e^{ct} = b \). Dividing on both sides by
\(\displaystyle e^{\frac{c(b+ \theta)}{a}}\) and factoring inside the exponential, we have
\(\displaystyle (at- \theta -b) e^{\frac{c}{a}(at- \theta -b)} = b e^{-\frac{c(b+ \theta)}{a}}\).
Now, multiply both sides by \(\displaystyle \frac{c}{a}\) and the equation becomes
\(\displaystyle \frac{c}{a}(at- \theta -b) e^{\frac{c}{a}(at- \theta -b)} = \frac{bc}{a} e^{-\frac{c(b+ \theta)}{a}}\).
The equation is now in the form \(\displaystyle Xe^X=y\), which implies that we can use the Lambert W function, defined
by \(\displaystyle \operatorname{W} \left (Xe^X \right ) = X\). Therefore, applying this function to both sides of the
equation, solving for t, and remembering that \(\displaystyle c= \frac{a}{b}\) gives
\(\displaystyle t= \frac{b}{a} \operatorname{W} \left ( e^{-(1+\frac{\theta}{b})} \right ) + \frac{\theta+b}{a}\).
We seek the value of R that minimizes this equation. The derivative of W is
\(\displaystyle \frac{\operatorname{W}(x)}{x(\operatorname{W}(x)+1)}\). Back-substituting for a and b, using this equation
and the chain rule, setting the result equal to zero, and clearing fractions gives
\(\displaystyle 2 I \omega_{f}^2 R^3 \left ( 1 + \operatorname{W} \left
(e^{-\left (1+\frac{\theta \varepsilon \tau_{s}}{I \omega_{f}^2 R^3} \right )}
\right ) \right )^2 +
\theta \varepsilon \tau_{s} \left (-1+ 2 \operatorname{W} \left
(e^{-\left (1+\frac{\theta \varepsilon \tau_{s}}{I \omega_{f}^2 R^3} \right )} \right ) \right )=0 \)
as the equation to be solved for R.
Do I have any errors?
\(\displaystyle \tau_{s}\) is the stall torque and \(\displaystyle \omega_{f}\) is the free angular speed. Our objective is to
find the value of the gear ratio \(\displaystyle R \) that minimizes the time required to reach a certain angular displacement
\(\displaystyle \theta \).
We begin from the equation \(\displaystyle \tau = I \frac{d \omega}{dt} \). Setting the two torque expressions equal and rearranging, we get
\(\displaystyle \frac{d \omega}{dt} + \frac{ \tau_{s} \omega}{I \omega_{f}} =
\frac{\tau_{s}}{I}\).
The integrating factor is \(\displaystyle \mu = e^{\int \frac{\tau_{s}}{I \omega_{f}} dt} = e^{\frac{\tau_{s} t}{I \omega_{f}}}
\). Multiplying all the terms of the equation by this factor and recognizing that the left side becomes a derivative, we obtain
\(\displaystyle \left ( \omega e^{\frac{\tau_{s} t}{I \omega_{f}}} \right )' = \frac{\tau_{s}}{I}
e^{\frac{\tau_{s} t}{I \omega_{f}}} \).
Integrating both sides from 0 to t and using the fact that \(\displaystyle \omega=0\) when \(\displaystyle t=0\), we obtain after rearrangement
\(\displaystyle \omega(t)= \omega_{f} \left (1-
e^{- \frac{\tau_{s} t}{I \omega_{f}}} \right )\).
Integrating again from 0 to t and using the fact that \(\displaystyle \theta=0\) when \(\displaystyle t=0\), we obtain
\(\displaystyle \theta(t)= t \omega_{f} - \frac{I \omega_{f}^2}{\tau_{s}}
\left (1 + e^{- \frac{\tau_{s} t}{I \omega_{f}}} \right)\).
Making the substitutions \(\displaystyle \tau_{s} \mapsto
\frac{\varepsilon \tau_{s}}{R}\) (where \(\displaystyle \varepsilon\) is efficiency) and
\(\displaystyle \omega_{f} \mapsto \omega_{f} R\), the equation becomes
\(\displaystyle \theta(t)= t \omega_{f} R - \frac{I \omega_{f}^2 R^3}{\varepsilon \tau_{s}}
\left (1 + e^{- \frac{\varepsilon \tau_{s} t}{I \omega_{f} R^2}} \right )\).
Next, we set
\(\displaystyle a= \omega_{f} R\),
\(\displaystyle b= \frac{I \omega_{f}^2 R^3}{\varepsilon \tau_{s}}\), and
\(\displaystyle c= \frac{a}{b}\).
The equation becomes \(\displaystyle \theta= at -b e^{-ct}-b\). We wish to invert this to get \(\displaystyle \theta\) as a function of t.
Adding b to both sides, multiplying by \(\displaystyle e^{ct}\), and combining the \(\displaystyle e^{ct}\) terms, we have
\(\displaystyle (at- \theta -b) e^{ct} = b \). Dividing on both sides by
\(\displaystyle e^{\frac{c(b+ \theta)}{a}}\) and factoring inside the exponential, we have
\(\displaystyle (at- \theta -b) e^{\frac{c}{a}(at- \theta -b)} = b e^{-\frac{c(b+ \theta)}{a}}\).
Now, multiply both sides by \(\displaystyle \frac{c}{a}\) and the equation becomes
\(\displaystyle \frac{c}{a}(at- \theta -b) e^{\frac{c}{a}(at- \theta -b)} = \frac{bc}{a} e^{-\frac{c(b+ \theta)}{a}}\).
The equation is now in the form \(\displaystyle Xe^X=y\), which implies that we can use the Lambert W function, defined
by \(\displaystyle \operatorname{W} \left (Xe^X \right ) = X\). Therefore, applying this function to both sides of the
equation, solving for t, and remembering that \(\displaystyle c= \frac{a}{b}\) gives
\(\displaystyle t= \frac{b}{a} \operatorname{W} \left ( e^{-(1+\frac{\theta}{b})} \right ) + \frac{\theta+b}{a}\).
We seek the value of R that minimizes this equation. The derivative of W is
\(\displaystyle \frac{\operatorname{W}(x)}{x(\operatorname{W}(x)+1)}\). Back-substituting for a and b, using this equation
and the chain rule, setting the result equal to zero, and clearing fractions gives
\(\displaystyle 2 I \omega_{f}^2 R^3 \left ( 1 + \operatorname{W} \left
(e^{-\left (1+\frac{\theta \varepsilon \tau_{s}}{I \omega_{f}^2 R^3} \right )}
\right ) \right )^2 +
\theta \varepsilon \tau_{s} \left (-1+ 2 \operatorname{W} \left
(e^{-\left (1+\frac{\theta \varepsilon \tau_{s}}{I \omega_{f}^2 R^3} \right )} \right ) \right )=0 \)
as the equation to be solved for R.
Do I have any errors?
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