- 01 Mar '17 03:29 / 3 editsI'm going to frame this question in context of an example.

A hoverboard and rider have a combined mass = m

It carries fuel mass = m_fuel

The relative velocity of the exhaust gases are constant = v_exh

A question might be posed: how long can it remain hovering ( a= 0,v=0)

Starting with Newtons 2nd Law: Σ**F**= d/dt ( m ***v**)

↑+ Σ**F**= -m * g - m_fuel * g + F_thrust = m * d**v**/dt + v * dm/dt

RHS:

m * d**v**/dt = 0 ( body is not accelerating, a= 0 )

v * dm/dt = 0 ( body is stationary, v=0)

So we have:

-m * g - m_fuel * g**+**F_thrust = 0 Eq1

(m+m_fuel) * g = F_thrust ( rearrange and dividng through by -1)

Here is where I'm looking for some clarification. The magnitude of the thrust force is given by: d/dt(m_fuel)*v_exh. The direction ( or sign ) seems to be opposite for Eq1 to be consistent. F_thrust =**-**d/dt(m_fuel)*v_exh.

Rewriting Eq1:

↑+ Σ**F**= -m * g - m_fuel * g**-**d/dt(m_fuel)*v_exh = 0

So why the negative? Some books say because the mass flow rate of fuel is decreasing d/dt(m_fuel), I can buy that, but I don't recall ever needing to flip signs in a typical force balance, or that is necessary to define (in advance) the direction of the rate? They typically define themselves in the act of performing the mechanics so long as a consistent coordinate system is used, correct?

What I'm personally gravitating to here is that "F_thrust" is actually a reaction force.

F_thrust + d/dt (m_fuel)*v = 0

Thus, when it is put in terms of d/dt(m_fuel)*v_exh the opposite direction must be assumed.

F_thrust = -d/dt (m_fuel)*v

So, is this a propper way to interpret this?

Performing a change of variables, separating, and integrating finishes out the example, nothing overly exciting there.

Thanks for any help - 01 Mar '17 12:06 / 1 edit

One thing I was thinking, it doesn't matter how the thrust comes about, it is still a matter of the energy stored either chemically or electrically. For instance, there has just been developed a real hoverboard with no rocket, strictly magnetic, the caveat is it cannot fly over a non conductor but on top of a steel plate for instance, the magnetic interaction allows real hovering and propulsion or at least hovering capability.*Originally posted by joe shmo***I'm going to frame this question in context of an example.**= d/dt ( m *

A hoverboard and rider have a combined mass = m

It carries fuel mass = m_fuel

The relative velocity of the exhaust gases are constant = v_exh

A question might be posed: how long can it remain hovering ( a= 0,v=0)

Starting with Newtons 2nd Law: Σ[b]F**v**)

↑+ Σ**F[ ...[text shortened]... and integrating finishes out the example, nothing overly exciting there.**

Thanks for any help

The gist of that is the force given by the magnetic field has to be exactly the same as the force involved in a physical expelling of fuel to give that thrust, the equations should be the same as far as the amount of energy it takes to keep the thing afloat, magnetic or chemical reactions aside, with the additional values of efficiency. Of course there would be no flow elements in the electromagnetic version except current flow which follows its own set of rules. - 02 Mar '17 02:20 / 1 edit

They are a little bit different scenarios. When using an electromagnet to hover, the power would be more or less constant. The weight of the system is not dependent on how much power is expended by the electromagnetic coil.*Originally posted by sonhouse***One thing I was thinking, it doesn't matter how the thrust comes about, it is still a matter of the energy stored either chemically or electrically. For instance, there has just been developed a real hoverboard with no rocket, strictly magnetic, the caveat is it cannot fly over a non conductor but on top of a steel plate for instance, the magnetic interacti ...[text shortened]... elements in the electromagnetic version except current flow which follows its own set of rules.**

However, In the case of a chemically propelled hoverboard, mass that the rider carries is being expelled through the nozzel as power is consumed. The weight of the rider and board are maintained, but the weight of carried fuel is not. As a result, as time goes on, progressively less power is required to hover. Until the fuel the rider is carrying runs out entirely. - 02 Mar '17 03:18

The minus sign is needed because v_exh is the*Originally posted by joe shmo***I'm going to frame this question in context of an example.**= d/dt ( m *

A hoverboard and rider have a combined mass = m

It carries fuel mass = m_fuel

The relative velocity of the exhaust gases are constant = v_exh

A question might be posed: how long can it remain hovering ( a= 0,v=0)

Starting with Newtons 2nd Law: Σ[b]F**v**)

↑+ Σ**F[ ...[text shortened]... and integrating finishes out the example, nothing overly exciting there.**

Thanks for any help*speed*and not the*velocity*of the exhaust gas. - 02 Mar '17 13:17

I would imagine the difference between fully loaded and empty of propellant wouldn't be that great, say you have a rider massing 100 Kg and you have 1 kg of fuel that lasts say 10 minutes. There would only be a 1% change in weight maximum. Of course I don't know jack about how much fuel it would take and how long it would last but it seems to me there would not be much difference between full and empty of fuel as to the weight of the entire system with rider onboard. Of course without rider, there would be a drastic % change fully fueled to empty weight.*Originally posted by joe shmo***They are a little bit different scenarios. When using an electromagnet to hover, the power would be more or less constant. The weight of the system is not dependent on how much power is expended by the electromagnetic coil.**

However, In the case of a chemically propelled hoverboard, mass that the rider carries is being expelled through the nozzel as pow ...[text shortened]... sively less power is required to hover. Until the fuel the rider is carrying runs out entirely. - 03 Mar '17 02:05

I don't know,...Maybe in this hoverboard scenario it can be neglected, but in large chemical rockets up to 90% of their mass is fuel. Obvioulsy it would be a large mistake to neglect the weight of fuel there.*Originally posted by sonhouse***I would imagine the difference between fully loaded and empty of propellant wouldn't be that great, say you have a rider massing 100 Kg and you have 1 kg of fuel that lasts say 10 minutes. There would only be a 1% change in weight maximum. Of course I don't know jack about how much fuel it would take and how long it would last but it seems to me there would ...[text shortened]... nboard. Of course without rider, there would be a drastic % change fully fueled to empty weight.** - 03 Mar '17 11:23

Don't forget, in large rockets you want several G's of acceleration whereas a hoverboard needs only 1. That triples the time it can hover or 1/3 the reaction mass.*Originally posted by joe shmo***I don't know,...Maybe in this hoverboard scenario it can be neglected, but in large chemical rockets up to 90% of their mass is fuel. Obvioulsy it would be a large mistake to neglect the weight of fuel there.** - 03 Mar '17 19:49

You are bucking a negative 1 g accel so you have to put in a positive 1 g of accel. If you were in outer space away from gravity wells you would indeed be undergoing 1 g of accel.*Originally posted by joe shmo***Actually, the hover board requires 0 g's of acceleration.**

So energy wise you can consider it a 1 g accel. - 06 Mar '17 15:25

Sorry to bring up side issues, your question is the math involved and I kind of sidetracked that.*Originally posted by sonhouse***You are bucking a negative 1 g accel so you have to put in a positive 1 g of accel. If you were in outer space away from gravity wells you would indeed be undergoing 1 g of accel.**

So energy wise you can consider it a 1 g accel. - 06 Mar '17 20:43 / 2 edits

Here is a problem to figure out: Like I said, a hoverboard based on rockets would indeed accel at 1 g away from our gravity well.*Originally posted by joe shmo***Its alright sonhouse. Over the years I've come to expect that from you. Your sidetracks are always welcome on my threads.**

But what about the hoverboard based on electro-magnets? It pushes up at one g but suppose you were in space away from our gravity well, say halfway to Mars or some such, and you had a very massive metal sheet that the hoverboard was bound to but pushing away from the sheet, say held by some string or other. Then you cut the string, how fast would you be doing say 10 seconds later. Obviously the thrust would go down quite severely, I would assume a lot more than inverse square law stuff.

You could assume a total weight on Earth of 100 Kg, passenger plus hoverboard electromagnets and associated electronics and power source, batteries? Nuclear? Solar, doesn't matter much for this problem. - 07 Mar '17 03:10 / 1 edit

Well, I don't think that is a very straightforward scenario, but lets start with some definitions.*Originally posted by sonhouse***Here is a problem to figure out: Like I said, a hoverboard based on rockets would indeed accel at 1 g away from our gravity well.**

But what about the hoverboard based on electro-magnets? It pushes up at one g but suppose you were in space away from our gravity well, say halfway to Mars or some such, and you had a very massive metal sheet that the hover ...[text shortened]... d electronics and power source, batteries? Nuclear? Solar, doesn't matter much for this problem.

The Electromagnet will be body "A"

The Flat Plate will be body "B"

The distance between the bodys "A" and "B" will be "L"

The electromagnetic pushing force will simply be "F(L)"

Examining body "A" (the electromagnetic hoverboard) using Newtons Second Law:

ΣF = m_A * d/dt ( v_A) = F(L) Eq1

Examining body "B" ( The plate) using Newtons Second Law:

ΣF = m_B * d/dt ( v_B) = - F(L) Eq2

Equating Eq1 and Eq2

m_A * d/dt ( v_A) = - m_B * d/dt ( v_B) Eq3

Now im going to randomly choose to put v_B in terms of v_A

From relative motion the velocity of B equals the velocity of A plus the relative velocity of B to A

v_B = v_A + v_B/A

Substitute that into Eq 3

m_A * d/dt ( v_A) = - m_B * d/dt ( v_A + v_B/A)

Then rearrange for v_A in terms of the relative velocity between B and A.

d/dt ( v_A ) = -m_B / (m_A + m_B ) * d/dt ( v_B/A )

Now, lets back substitute this expression into Eq 1

m_A * - m_B / (m_A + m_B ) * d/dt ( v_B/A ) = F(L)

Now, lets get the relative velocity in terms of our distance parameter "L"

d/dt ( v_B/A ) = d²/dt² ( L )

Substituting that bit in we have:

m_B * m_A / (m_A + m_B ) * d²/dt² ( L ) + F(L) = 0

Now, as you said the magnetic force follows some inverse law in relation to our distance "L". which is fairly complicated and fuzzy itself. Furthermore, unfortunately ( for me ) that makes this a Second Order Non-Linear Differential Equation which we need to solve. ( If I did that analysis properly). Which basically means I'm out of my depth ( or perhaps have been out of my depth this whole explanation ). Maybe someone can take it from here or restate the problem in terms of energy somehow to simplify it. - 07 Mar '17 13:11

This seems to boil down to exactly what level of inverse squareness is involved. Maybe it's an inverse cube law. Not sure.*Originally posted by joe shmo***Well, I don't think that is a very straightforward scenario, but lets start with some definitions.**

The Electromagnet will be body "A"

The Flat Plate will be body "B"

The distance between the bodys "A" and "B" will be "L"

The electromagnetic pushing force will simply be "F(L)"

Examining body "A" (the electromagnetic hoverboard) using Newtons Second ...[text shortened]... someone can take it from here or restate the problem in terms of energy somehow to simplify it.