Formula derivation connecting vertical water flowrate & horizontal distance moved by a suspended sphere

[printereater]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$
$$\dot{m}=\frac{T\sin(\beta )}{v_{0}\sin(\theta)}\rightarrow (1)$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=\dot{m}(v_{i}-v_{0}\cos(\theta))$$
$$\dot{m}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}\rightarrow (2)$$

$$(1)=(2)$$
$$\therefore \frac{T\sin(\beta )}{v_{0}\sin(\theta)}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}$$

 

[erobz]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$
$$\dot{m}=\frac{T\sin(\beta )}{v_{0}\sin(\theta)}\rightarrow (1)$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=\dot{m}(v_{i}-v_{0}\cos(\theta))$$
$$\dot{m}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}\rightarrow (2)$$

$$(1)=(2)$$
$$\therefore \frac{T\sin(\beta )}{v_{0}\sin(\theta)}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}$$

You aren't getting what I'm saying ( or I don't understand what making $\dot m$ "the subject " is). Either way, you want each equation to not have any velocity parameters. Just the force terms, mass flowrate $\dot m$, relevant area $A_i,A_o$, and known physical properties (density). Don't join the equations yet.

Think about equating $\dot m = \dot m_i = \dot m_o$ to eliminate each velocity $v_i,v_o$ from each equation.

 

[printereater]

ohh so I need to put $v_{o}$ and $v_{i}$ in terms of $\dot{m}$,$\rho$ and $A$ using $\dot{m}=\rho vA$

 

[erobz]

ohh so I need to put $v_{o}$ and $v_{i}$ in terms of $\dot{m}$,$\rho$ and $A$ using $\dot{m}=\rho vA$

yeah for each location ( inlet/outlet) that holds.

 

[printereater]

Got it. It's about to be 1:30 am now, I will head to bed and continue tomorrow. Thank you:)

 

[erobz]

Got it. It's about to be 1:30 am now, I will head to bed and continue tomorrow. Thank you:)

Get some rest, I'll be unavailable for the weekend. If you want to continue, feel free, someone will help you. If you want to wait till Monday, that is fine too.

 

[printereater]

Oh no. If that's the case, i don't mind sacrificing my sleep HAHAHAH let's continue. Am I close to the final equation

 

[erobz]

You're, going to get tired, and learning abilities are going to decline, and mistakes are going to start to happen in your math. There is certainly no way we can get through the Reynolds Transport Theorem - "The Momentum Equation", with any level of understanding so quick.

Your goal (for the weekend- if waiting for me at least - which is not necessary ) after eliminating the velocities, is then to eliminate the Tension force from the equations by substitution, and then solve the resulting equation containing only parameters $\dot m, A_i, A_o,\rho, \theta$ for the angle the ball makes w.r.t vertical $\beta$.

I strongly suggest you just take a rest, and work at it refreshed.

 

[printereater]

yea you are right, I will just continue tomorrow. Thank you so much, once again

 

[printereater]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{o}\sin(\theta))\rightarrow (1)$$
$$\dot{m}=\rho v_{o}A_{o}$$
$$v_{o}=\frac{\dot{m}}{\rho A_{o}}\rightarrow (2)$$
Substitute (2) into (1) :
$$T\sin(\beta )=\dot{m}(\frac{\dot{m}}{\rho A_{o}})\sin(\theta)$$
$$\therefore T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$

In y-coordinate direction
$$T\cos(\beta )-W=-\dot{m}v_{o}\cos(\theta))+mv_{i}\rightarrow (3)$$
$$v_{i}=\frac{\dot{m}}{\rho A_{i}}\rightarrow (4)$$
Substitute (2) and (4) into (3):
$$T\cos(\beta )-W=-\dot{m}\frac{\dot{m}}{\rho A_{o}}\cos(\theta)+m\frac{\dot{m}}{\rho A_{i}}$$
$$\therefore T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

 

[printereater]

In x-coordinate direction,
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}}{\rho A_{i}A_{o}}-\frac{\dot{m}^{2}A_{i}\cos(\theta)}{\rho A_{i}A_{o}}+\frac{W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}\rightarrow (6)$$

(5)=(6),
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{ A_{i}\cos(\beta )}$$

 

[erobz]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{o}\sin(\theta))\rightarrow (1)$$
$$\dot{m}=\rho v_{o}A_{o}$$
$$v_{o}=\frac{\dot{m}}{\rho A_{o}}\rightarrow (2)$$
Substitute (2) into (1) :
$$T\sin(\beta )=\dot{m}(\frac{\dot{m}}{\rho A_{o}})\sin(\theta)$$
$$\therefore T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$

In y-coordinate direction
$$T\cos(\beta )-W=-\dot{m}v_{o}\cos(\theta))+mv_{i}\rightarrow (3)$$
$$v_{i}=\frac{\dot{m}}{\rho A_{i}}\rightarrow (4)$$
Substitute (2) and (4) into (3):
$$T\cos(\beta )-W=-\dot{m}\frac{\dot{m}}{\rho A_{o}}\cos(\theta)+m\frac{\dot{m}}{\rho A_{i}}$$
$$\therefore T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

Keep factoring the y coordinate equation to clean it up. Otherwise, ok. Then keep going.

 

[erobz]

In x-coordinate direction,
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}}{\rho A_{i}A_{o}}-\frac{\dot{m}^{2}A_{i}\cos(\theta)}{\rho A_{i}A_{o}}+\frac{W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}\rightarrow (6)$$

(5)=(6),
$$\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}$$
$$\frac{\dot{m}^{2}\sin\(\theta)}{\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{ A_{i}\cos(\beta )}$$

Your algebra looks unnecessarily complicated here. Just use the x direction equation to eliminate $T$ in the y direction.

 

[printereater]

Oh alright

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

 

[erobz]

$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

factor the RHS a bit more and then proceed to solve for $\beta$, and you are done with that part.

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})+W$$
$$\frac{\rho A_{o}\tan(\beta )}{\dot{m}^{2}\sin(\theta)}=\frac{\rho}{\dot{m}^{2}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{1}{W}$$
$$\tan(\beta )=\frac{\rho\dot{m}^{2}\sin(\theta)}{\dot{m}^{2}\rho A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}$$
$$\tan(\beta )=\frac{\sin(\theta)}{A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}$$
$$
\tan(\beta )=\arctan[\frac{\sin(\theta)}{A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}]$$

 

[printereater]

@erobz @Delta2 Please take a look at it when you are free and let me know if I am on the right track.

 

[erobz]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})+W$$

You go off the rails on the next step. In general, $\frac{1}{a+b} \neq \frac{1}{a}+\frac{1}{b}$.

 

[printereater]

You go off the rails on the next step. In general, $\frac{1}{a+b} \neq \frac{1}{a}+\frac{1}{b}$.

Sorry, I just saw your message. I will try again today. The goal is it to factorise the right hand side and make $\beta$ the subject of the equation right

 

[erobz]

Sorry, I just saw your message. I will try again today. The goal is it to factorise the right hand side and make $\beta$ the subject of the equation right

You are trying to solve for $\beta$ as a function of all the other variables you intend to measure. If that’s is what “making it the subject is “, then sure. You made an algebraic mistake when inverting both sides.

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W\rho A_{o}$$
$$\frac{1}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\dot{m}^{2}\rho\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}$$
$$\tan(\beta )=\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}$$
$$\beta =\arctan[\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}]$$ I am not sure if there is a nicer way to express $\beta$

 

[erobz]

I am not sure if there is a nicer way to express $\beta$

Its good enough for now. Do you understand how to use it?

Latex tip. Use automatically sized parenthesizes

Instead of $$ ( 1 + \frac{1}{x} ) $$

use code

$$ \left( 1 + \frac{1}{x} \right) $$

to get

$$ \left( 1 + \frac{1}{x} \right) $$

 

[printereater]

Hi yes I understand it. How can I use it to link the distance moved by the sphere and the flow rate though

 

[erobz]

Hi yes I understand it. How can I use it to link the distance moved by the sphere and the flow rate though

$\beta$ is the steady state angle, you need all the other variables measured (or controlled) and the equation spits out the angle the string makes with vertical (getting how far it moved is just trig), you can determine how well that fits with the angle $\beta_{exp}$ you experimentally measure. There is going to be error in all the variables to account for etc...

 

[printereater]

$\beta$ is the steady state angle, you need all the other variables measured (or controlled) and the equation spits out the angle the string makes with vertical (getting how far it moved is just trig), you can determine how well that fits with the angle you measure. There is going to be error in all the variables to account for etc...

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

 

[printereater]

There is going to be error in all the variables to account for etc...

Yes! Can you guide me through the calculations of uncertainties too please

 

[printereater]

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

I just realised, there are 4 variables that are changing each time based on the flow rate of water $\ A_{o},\ A_{i}, \theta, \dot{m}$. So I have no choice but to plot D against $\beta$ right

 

[erobz]

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

This is the steady state solution. $\dot m$ must be controlled. The idea is that if you control the flowrate to be constant, all other experimentally measured variables $\theta, A_i,A_o$ are constant too (for each flowrate). So, the equation predicts the angle of the string $\beta$. Then, you change the flow rate(but keep it steady), and re-measure the variables, and again the equation predicts the angle. You compare how well the equation matches what you experimentally measure for $\beta$ at each flowrate.

The goal is to see how well our theory about reality, matches reality. Will it be a decent model given our assumptions, and our inability to precisely measure the variables that contrive it? That is the purpose of an experiment.

 

[printereater]

Right, I get what you mean but for the graph I plot, am I going to plot $\beta$ against $\dot{m}$ or do I just plot $D=L\tan (\beta )$ to show the relationship between flowrate of water and sphere. With this formula I can show how $\beta$ changes with D to indirectly show the relationship between flowrate and D because what's affecting $\beta$ is $\dot{m}$ but I am wondering if there is a direct way to show the relationship flowrate and D.

 

[erobz]

Right, I get what you mean but for the graph I plot, am I going to plot $\beta$ against $\dot{m}$

Just do this. you are introducing more sources of error by adding new variables that need measured.

but I am wondering if there is a direct way to show the relationship flowrate and D.

Do you mean without measuring $\theta, A_i, A_o$?

 

[printereater]

Just do this. you are introducing more sources of error by adding new variables that need measured.

But then i would not be answering my research question though:(

Do you mean without measuring $\theta, A_i, A_o$?

I mean ideally I would love to have an equation in which I have D on the left hand side and $\dot{m}$ on the right hand side to directly express a relationship between these 2 variables.

 

[erobz]

I mean ideally I would love to have an equation in which I have D on the left hand side and $\dot{m}$ on the right hand side to directly express a relationship between these 2 variables.

A closed form solution of $\beta( \dot m )$...for all I know that could be a Nobel Prize in Physics.

 

[printereater]

oh Let's not go there then

 

[erobz]

oh Let's not go there then

By all means, give it shot if you are unsatisfied.