Energy Efficiency of Compressed-Air-Powered Parabola Riding Vehicle?

[metastable]

The equations seem to give absurd results with very high passenger vehicle masses:

 

[sysprog]

Obviously we can't see on your spreadsheet the equation or inequality expression that resulted in the apparently absurd result -- in addition to your having the ability to use $\LaTeX$ here to show your math readably, you also can use BBcode [code ] tags to show your code -- maybe you would care to show, cell by cell, your .xls code that generated the 'absurd' result $\dots$

 

[Dale]

Excel code is almost impossible to debug.

 

[Vanadium 50]

Excel code is almost impossible to debug.

A wall-O-numbers in Excel is still a wall-O-numbers.

 

[metastable]

I think its an error in excel rather than the formulas because it jumps from a reasonable answer when I keep adding zeros to suddenly a ridiculous answer. In other words, with one less $0$ at the end of the passenger vehicle mass I think it gives the correct answer still.

Suppose we have a $10^4$ $kg$ passenger vehicle with a $20$ $kg$ trailer containing $200$ $kg$ of water, perched at the top of a vertical $30$ $second$ drop in a vacuum, with a curved section of track at the bottom leading to a flat section that leads to the destination where there is a second curved section leading to a second vertical section…

At the bottom of the ramp, the passenger vehicle exerts a mechanical impulse equivalent to $9.54$ $MJ$, pushing off the $294$ $m/s$ trailer instead of the ground, which brings the trailer to a halt on the tracks…

^Assuming the passenger vehicle is using maglev in vacuum, and therefore does no work while coasting, the passenger vehicle is traveling $294$ $m/s$ in a straight line at the bottom of the ramp after the mechanical impulse. Assuming regen braking of the passenger vehicle at its destination (after climbing a second vertical ramp to the surface) results in $70$ % kinetic to kinetic conversion efficiency, and that only the empty trailer is lifted, a factor of $130$ % as much usable energy is recovered from the vehicle after it climbs to the surface than was exerted in the mechanical impulse at the bottom of the ramp, a net surplus if the geothermal energy required to evaporate this water is ignored. The surplus comes from the lowered gravitational potential energy of the $200$ $kg$ water at the bottom of the tunnel. If the full trailer is lifted with the regen braking energy instead of geothermal, then only $40$ % of the mechanical impulse energy is recoverable. If the full trailer is lifted from the recovered regen braking energy, then a total of $5.71$ $MJ$ energy was “unrecoverably” consumed accelerating and decelerating the $10^4$ $kg$ vehicle to $294$ $m/s$. If the $200$ $kg$ water is left at the bottom of the tunnel and allowed to evaporate naturally, then an “excess” of $2.88$ $MJ$ energy above and beyond the total mechanical impulse energy is obtained from regen braking the vehicle at $70$ % conversion efficiency at the destination, even after using some of the recovered energy is used to lift the empty trailer. The kinetic energy of the passenger vehicle after the mechanical impulse at the bottom of the ramp is a factor of $~760x$ greater than the total energy “unrecoverably” consumed when lifting the full (not empty) trailer all the way back to the surface, using the recovered kinetic energy via the regen braking after the journey.

 

[metastable]

First I will share an example of the equations I would use to calculate the power consumption of a given standard land vehicle at a given speed in a given set of conditions.

The ACME Corporation cannot guarantee the safety of a rider aboard an electric skateboard operated at a speed exceeding 100 mph100 mph100\ mph.

for comparison I will share a photo of, what I personally consider, a "standard car"-- quite literally the drivetrain of the fastest electric skateboard in the world, capable of 70+ $mph$:

 

[sysprog]

for comparison I will share a photo of, what I personally consider, a "standard car"-- quite literally the drivetrain of the fastest electric skateboard in the world, capable of 70+ $mph$:

View attachment 246979

Please sign your organ donor card before riding that thing, kid; how do you deal with the snap (aka jounce)? Oh, and, I have a friend whose pet kitty needs a cornea transplant. On a slightly more technical note, umm -- WB cartoon speed was measured in frames per second (12 fps doubled to 24); not really mph -- Mr. Coyote's speed is forever less than that of the Road Runner, speed in this instance being considered as the derivative of distance with respect to time -- the reasoning behind the standard for the time required to complete 1 frame (not for the artist; for the viewer) is hinted at by the following inequality constraint: $1(frametime) < 1(blinkofaneye)$. Actually, it has more to do with human brain visual image processing time -- the frame rate is determined by how many frames per second are required to cause the visual system to report continuous movement instead of separate still images.

And I didn't mean to hijack your thread.

@metastable posted me, and I mostly begged off for now, regarding his to-me-somewhat-quizzical spreadsheet code. My at-a-glance response was that the source of the problem may/might be in his having exhausted the capacity of the sqrt function.

 

[metastable]

@metastable posted me, and I mostly begged off for now, regarding his to-me-somewhat-quizzical spreadsheet code. My at-a-glance response was that the source of the problem may/might be in his having exhausted the capacity of the sqrt function.

For me, the most interesting part of the whole exercise was determining the formula for calculating how many Joules of Mechanical Impulse between the vehicle and the trailer will bring the trailer to a complete halt at the bottom of the ramp every time, depending on the Trailer Mass, Vehicle Mass, Free Fall Duration, Initial Velocity Before Free Fall, and the Gravitational Acceleration at Earth's Surface:

$W_i$ = $12.8*10^6$ = 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 ($m/s^2$)

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

 

[sysprog]

For me, the most interesting part of the whole exercise was determining the formula for calculating how many Joules of Mechanical Impulse between the vehicle and the trailer will bring the trailer to a complete halt at the bottom of the ramp every time, depending on the Trailer Mass, Vehicle Mass, Free Fall Duration, Initial Velocity Before Free Fall, and the Gravitational Acceleration at Earth's Surface:

$W_i$ = $12.8*10^6$ = 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 ($m/s^2$)

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

Please disambiguate the last-line expression -- in particular, the '/2' -- does that fractionate the entirety? -- it's ok to use $\frac 1 2$ and/or extra parens (instead of making me have to go with PEMDAS and the rest of my feeble memory of the order of operations) to keep it clear what you mean $\dots$

 

[metastable]

Please disambiguate the last-line expression -- in particular, the '/2' -- does that fractionate the entirety? -- it's ok to use 1212\frac 1 2 and/or extra parens (instead of making me have to go with PEMDAS and the rest of my feeble memory of the order of operations) to keep it clear what you mean

I'm not sure what you mean exactly, does this help?

240*0*30*9.80665+(240*30^2*9.80665^2+240*0^2)/(2)+(240^2*(0+30*9.80665)^2)/(2*1000)

 

[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]

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

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]

$M_{tot}V_b=M_vV_v$

^I haven't gone through your whole analysis yet but I had a question about this line... shouldn't it be:

$M_{trailer}V_b=M_vV_v$

 

[jbriggs444]

^I haven't gone through your whole analysis yet but I had a question about this line... shouldn't it be:

$M_{trailer}V_b=M_vV_v$

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]

Effectively what I used was this:

$m_1v_1=m_2v_2$

^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?

 

[jbriggs444]

Effectively what I used was this:

$m_1v_1=m_2v_2$

^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]

^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]

Effectively what I used was this:

$m_1v_1=m_2v_2$

^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]

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]

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:

$m_1v_1=m_2v_2$

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

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

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

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

 

[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:
$m_1v_1=m_2v_2$

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

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

I forgot to add:

$a = (v_f - v_i) / Δt$

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: