# B Land Based Oberth Manuever?

#### sysprog

It does help -- you've given me something to think about -- thanks for disambiguating $\dots$

#### metastable

Calling it the WORM, short for Water Oberth Ramp Mover seems appropriate, and the ramp / station terminals -- the wormhole.

#### jbriggs444

Homework Helper
$W_i=M_tV_iT_dG_e+(M_tT_d^2G_e^2+M_tV_i^2)/2+(M_t^2(V_i+T_dG_e)^2)/(2M_v)$
Let me be certain that my understanding is clear. We are talking about a trailer full of water behind a passenger vehicle. The water is not ejected in this scenario. It remains with the trailer throughout. There are no frictional losses. The whole assembly drops and gains kinetic energy. The passenger vehicle is then ejected at a velocity such that conservation of momentum brings the trailer to a stop.

If solving this from scratch, the relevant input parameters would seem to be the mass ratio of vehicle to trailer and the assembly velocity at the bottom of the tunnel. The other input parameters are just window dressing from which those two can be calculated.

The formula that is supplied above seems inordinately complex for such a simple situation. It is also written down without justification. Let us attack the problem with algebra. While I am not happy with all of the case conventions or the subscripts on the variable names, but I will continue in that style, re-using your variable names as much as possible.

I am going to skip past the (easy) calculation of velocity at the bottom and assume that as a given. Call it $V_b$. Now we apply a conservation of momentum argument to determine how fast the passenger vehicle must be ejected if the remaining momentum of the trailer becomes zero.

Let $V_v$ denote the passenger vehicle ejection velocity and $M_{tot}$ denote the combined mass of vehicle plus trailer.
$$M_{tot}V_b=M_vV_v$$
$$V_v=V_b\frac{M_{tot}}{M_v}$$
Or, easier yet, let $R$ denote the mass ratio, total mass to vehicle mass.
$$V_v = R V_b$$
We are after an energy measure. How much energy did it take to produce this ejection velocity? Let $E$ denote the energy delta between the final state (passenger vehicle ejected, trailer at rest) and the initial state (vehicle plus trailer moving at V_b).
$$E=\frac{1}{2}M_vV_v^2-\frac{1}{2}(M_{tot})V_b^2$$
Let's substitute in for $V_v$ using our formula in terms of $V_b$
$$E=\frac{1}{2}M_vR^2V_b^2-\frac{1}{2}M_{tot}V_b^2$$
Let's substitute one of the factors of R in that first term.
$$E=\frac{1}{2}M_{tot}RV_b^2-\frac{1}{2}M_{tot}V_b^2$$
That $\frac{1}{2}M_{tot}V_b^2$ term looks exactly like "energy at the bottom". So let's call it that.
$$E=E_{bottom}(R-1)$$
We can simplify this further.
$$R-1=\frac{M_{trailer}+M_v}{M_v}-1=\frac{M_{trailer}}{M_v}+\frac{M_v}{M_v}-1=\frac{M_{trailer}}{M_v}$$
So let's just use lower case $r$ to denote the trailer to vehicle mass ratio.
$$E=rE_{bottom}$$

There are easier ways to this simple result.

#### sysprog

Wow, @jbriggs444, (thank you for) a very extremely good post. $\uparrow$

#### metastable

$M_{tot}V_b=M_vV_v$

$M_{trailer}V_b=M_vV_v$

#### jbriggs444

Homework Helper

$M_{trailer}V_b=M_vV_v$
No. It was correct as written. Momentum before = momentum after. The momentum before is the total mass times the velocity of the complete assembly.

#### metastable

$m_1v_1=m_2v_2$

^If the mass of the trailer acquires the ramp-bottom velocity in an energetic interaction / push-off with the passenger section (bringing the trailer to a halt), then the passenger vehicle velocity is $v_2$ + $V_{bottom}$

The energy of the interaction between trailer and passenger section would be:

$W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2+V_b)^2$

Was it wrong?

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#### jbriggs444

Homework Helper

$m_1v_1=m_2v_2$

^If the mass of the trailer acquires the ramp-bottom velocity in an energetic interaction / push-off with the passenger section (bringing the trailer to a halt), then the passenger vehicle velocity is $v_2$ + $V_{bottom}$

The energy of the interaction between trailer and tank would be:

$W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2+V_b)^2$

Was it wrong?
As I understand this, you want $V_b$ to be the forward velocity of the trailer+vehicle assembly.

If we take $v_2$ to be the forward velocity of the vehicle relative to the pre-ejection assembly then the second term in your formula correctly reflects the final kinetic energy of the vehicle in the ground frame. But then the first term does not correctly reflect the final kinetic energy of the trailer in the ground frame. In addition, we are left with no deduction for initial kinetic energy in the ground frame.

It is difficult to second guess someone else's reasoning process, but you might have determined the initial kinetic energy of the assembly in the moving frame, the final kinetic energy of the trailer in the moving frame and the final kinetic energy of the vehicle in the ground frame. That does not work.

Kinetic energy is frame-dependent. You cannot mix and match picking one energy from frame A and another energy from frame B and expect them to add coherently.

#### metastable

$W_i$ = $12.8∗10^6$ = Mechanical Impulse Energy ($Joules$) That Brings Trailer to 0 Velocity At Ramp Bottom
$M_t$ = $240$ = Trailer Tank + Water Mass ($Kilograms$)
$M_v$ = $10^3$ = Passenger Vehicle Mass ($Kilograms$)
$T_d$ = $30$ = Free Fall Duration ($Seconds$)
$V_i$ = $0$ = Initial Velocity Before Free Fall ($m/s$)
$G_e$ = $9.80$ = Gravity Acceleration at Earth’s Surface ($m/s^2$)
^Using your separate method do you calculate the same value for $W_i$? The reason I ask is I don't understand part of your method at the beginning.

#### jbriggs444

Homework Helper
^Using your separate method do you calculate the same value for $W_i$? The reason I ask is I don't understand part of your method at the beginning.
I am not willing to run the numbers if you are not willing to do the algebra.

#### metastable

$m_1v_1=m_2v_2$

^If the mass of the trailer acquires the ramp-bottom velocity in an energetic interaction / push-off with the passenger section (bringing the trailer to a halt), then the passenger vehicle velocity is v2v2v_2 + VbottomVbottomV_{bottom}

The energy of the interaction between trailer and passenger section would be:

$W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2+V_b)^2$

Was it wrong?
In my above calculation, $v_2$ was intended to be the velocity of the passenger section, relative to the $V_{bottom}$ rest frame of the trailer+vehicle at the bottom of the tunnel before the push off, so $v_2 + V_{bottom}$ is the final velocity of the passenger vehicle while still on the flat section at the bottom of the tunnel.

#### jbriggs444

Homework Helper
In my above calculation, $v_2$ was intended to be the velocity of the passenger section, relative to the $V_{bottom}$ rest frame of the trailer+vehicle at the bottom of the tunnel before the push off, so $v_2 + V_{bottom}$ is the final velocity of the passenger vehicle while still on the flat section at the bottom of the tunnel.
Yes. As I explained, you need to pick a frame of reference for your energy balance.

#### metastable

The energy of the interaction between trailer and passenger section would be:

$W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2+V_b)^2$

Was it wrong?
I misspoke here, I meant:

$W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2)^2$

but the final velocity of the passenger vehicle relative to the ground is:

$v_2+V_{bottom}$

#### metastable

and therefore the final kinetic energy of the passenger vehicle relative to the ground at the bottom of the tunnel is:

$(1/2)m_2(v_2+V_{bottom})^2$

#### jbriggs444

Homework Helper
I misspoke here, I meant:

$W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2)^2$
Yes, that would be a correct calculation for incremental work done.

#### metastable

bringing all the equations together, we have:

$m_1v_1=m_2v_2$

$v_2=(m_{trailer}V_{bottom})/m_{passenger}$

$W_i=(1/2)m_{trailer}V_{bottom}^2+(1/2)m_{passenger}(v_2)^2$

$V_{passenger}=v_2+V_{bottom}$

$W_{KEpassenger}=(1/2)m_{passenger}(v_2+V_{bottom})^2$

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#### sysprog

bringing all the equations together, we have:

$m_1v_1=m_2v_2$

$v_2=(m_{trailer}V_{bottom})/m_{passenger}$

$W_i=(1/2)m_{trailer}V_{bottom}^2+(1/2)m_{passenger}(v_2)^2$

$V_{passenger}=v_2+V_{bottom}$

$W_{KEpassenger}=(1/2)m_{passenger}(v_2+V_{bottom})^2$
Thanks for that.

#### metastable

bringing all the equations together, we have:
$m_1v_1=m_2v_2$

$v_2=(m_{trailer}V_{bottom})/m_{passenger}$

$W_i=(1/2)m_{trailer}V_{bottom}^2+(1/2)m_{passenger}(v_2)^2$

$a = (v_f - v_i) / Δt$

where:

$a$ is the acceleration, $v_i$ initial velocity, $v_f$ final velocity, $Δt$ is the acceleration time

combining, rearranging & simplifying all the equations:

$a = (V_{bottom} - v_i) / Δt$
$m_1v_1=m_2v_2$
$v_2=(m_{trailer}V_{bottom})/m_{passenger}$
$W_i=(1/2)m_{trailer}V_{bottom}^2+(1/2)m_{passenger}(v_2)^2$

we get:

$W_i=M_tV_iT_dG_e+(M_tT_d^2G_e^2+M_tV_i^2)/2+(M_t^2(V_i+T_dG_e)^2)/(2M_v)$

$W_i$ = $1.28*10^7$ = Mechanical Impulse Energy ($Joules$) That Brings Trailer to 0 Velocity At Ramp Bottom
$M_t$ = $240$ = Trailer Tank + Water Mass ($Kilograms$)
$M_v$ = $10^3$ = Passenger Vehicle Mass ($Kilograms$)
$T_d$ = $30$ = Free Fall Duration ($Seconds$)
$V_i$ = $0$ = Initial Velocity Before Free Fall ($m/s$)
$G_e$ = $9.80$ = Gravity Acceleration at Earth’s Surface ($m/s^2$)

#### sysprog

Did you by this:

$W_i=M_tV_iT_dG_e+(M_tT_d^2G_e^2+M_tV_i^2)/2+{{(M_t^2(V_i+T_dG_e)^2)/(2M_v)}}$

mean this:

$$W_i=\frac {M_tV_iT_dG_e+(M_tT_d^2G_e^2+M_tV_i^2)}{2+{{(M_t^2(V_i+T_dG_e)^2)/(2M_v)}}}$$

???

#### metastable

I don’t believe so because when I input it in a calculator it looks like this:

"Land Based Oberth Manuever?"

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