Relative size of combatants

Are we discussing just definitions here?

No work being done, but energy is being expended to keep the object levitating (countering gravity) assuming that gravity is still in effect. Force but no displacement.

Now if you are saying that the spell uses the energy to counter gravity, so there is no additional energy required to keep the levitating object levitating, then I understand that well enough.

Louis R wrote:Incorrect. As you should be aware simply from looking around you.

Assuming, of course, that you didn't write that immediately after jumping out of a window. In which case you will have by now hit the ground and probably won't be responding for a while. On the way there, though, you would have noticed movement, although whether you would have ascribed that movement to yourself or your surroundings isn't evident, and you would indeed have been accelerating at 32 feet per second per second. Nothing around you would be, though, which is why reaching the ground is so... awkward.

Should you choose to expire in a more static fashion, we can stick your neck in a noose, haul you to the yardarm and pay off the line around the nearest belaying pin. You would be very much in the way should we need to trim the sails, but I can assure you that no power will be needed to keep you up there. Interestingly, should we pay the line through an electric winch, and try to use the motor to hold you up, the power needed, for the short time it took for the motor to reduce itself to slag, would actually be a very large multiple [20-50x] of the power used to get you up there in the first place. But that is due to the characteristics of an electric motor with a locked rotor, not because we need any power to keep you dangling.

Which brings us around to levitation spells again. Any magical power - whatever that means - consumed in the steady state is going to be dissipated in the internal workings of the spell, because no work is being done on the load. And, judging by the physical power that wasn't needed to raise the load in the first place, that amount isn't all that closely related to the size of the load.

Your definition of work might be magically correct, but is not consistent with the definition of work used in our physics. The work done by my floor to hold me up is, well, zero, and the floor of my den does not get tired or run out of energy holding me up.

Work = \integral_start^Finish F(x) dx (for 1-Dimensional motion)

For levitation the displacement parallel to F is zero, so the work is zero.

George
D.Sc., MIT (Physics) '73.

PeterZ wrote:There is movement. An object in Earth's gravity well accelerates at 32 feet/sec/sec. An object at rest will begin to accelerate towards the Earth's center unless another force acts on it. To levitate at sea level, sufficient force to accelerate the object at 32 feet/sec/sec away from Earth's center of mass is required. This assumes nothing else supports the object. Work can be calculated as the force necessary to accelerate an object 32 feet/sec/sec by the distance the object will travel at that acceleration for the length of time the object is made to levitate. Because baring any force to counter gravity, an object will accelerate towards the Earth's center of mass.

Temporarily changing the local gravity might take a lot or a very little arcane energy. Only RFC knows.

bkwormlisa wrote:I agree that holding up something against gravity (or negating gravity for a given mass, or whatever) should be very power intensive, in the sense that it should drain the spell accumulator quickly. And they can only hold up a dragon for a short period of time, which is highly suggestive.

However, what Louis R actually said is that no power is expended while holding something still. In physics, the definition of power is work per time, and the definition of work is that of a force times a displacement - there must be movement, or no work is done. So technically, to a physicist, holding something hovering in midair does no work and therefore takes no power. (I tried to calculate how much energy it takes to hold something up with telekinesis once, and there's no easy answer.) So Louis's statement that is takes no power is true, from a certain point of view (depending on the specific definition of power). My guess is that that's what he meant.

I'm surprised that they do hover sliders, really. Shaylar told us they have to keep the cars light due to the levitation, so why don't they have tracks and just use magic for propulsion? Granted, it would be more limited and take more time to set up, plus you'd have to build a slider track each way instead of only one (two sliders meeting head-on just slide around each other and keep going, as it is), but surely it would still be more efficient, especially for cargo and heavy transport. I wonder if they just never thought of it and will now that they've seen a railroad? Or is it a limitation because of the amount of steel it would take, or the labor required?

I understand now. Thank you.

I still have to ask how much energy is used to prevent a body not supported by anything physical from accelerating at 32 feet/sec/sec? Assuming gravity is not altered in any way and energy/force is required to do the levitating.

Wouldn't that be the amount of energy required to perform the work of lifting that body at a rate of 32 feet/sec/sec for however long the spell is in effect for? After all, absent the supporting energy the body would accelerate at 32 feet/sec^2. The default state here is acceleration and to counter that state would require and equal and opposite force. As I recall Newton's First Law of Motion pretty much covers this.

All this goes out the window if arcane energy is used to change gravity so that it no longer pulls on the body.

phillies wrote:Your definition of work might be magically correct, but is not consistent with the definition of work used in our physics. The work done by my floor to hold me up is, well, zero, and the floor of my den does not get tired or run out of energy holding me up.

Work = \integral_start^Finish F(x) dx (for 1-Dimensional motion)

For levitation the displacement parallel to F is zero, so the work is zero.

George
D.Sc., MIT (Physics) '73.

phillies wrote:Your definition of work might be magically correct, but is not consistent with the definition of work used in our physics. The work done by my floor to hold me up is, well, zero, and the floor of my den does not get tired or run out of energy holding me up.

Work = \integral_start^Finish F(x) dx (for 1-Dimensional motion)

For levitation the displacement parallel to F is zero, so the work is zero.

George
D.Sc., MIT (Physics) '73.

But this is almost a definitional issue. A 5 pound weight sitting on the floor requires no work from the floor, it just imparts a static load. Similarly a 5 pound weight suspended by a rope requires no work from the rope it just imparts a static force along the rope.

But if I hold the 5 pound weight at arms length, I'm still doing no physics work, but the muscles in my arm are consuming chemical energy to keep the weight from descending to towards the floor. (If I let it drop to my side then I'd only need enough chemical energy to keep a grip and the rest of the force could be statically handled by the bones and connecting tissues of my arm). Similarly if I used a an RC helicopter or small rocket to hover that 5 pound load several feet off the ground they're going no physics work, but they're consuming additional chemical (or electrical) energy to keep the payload weight from moving. Far more than they'd need to just hold themselves steady.

So sure, it takes no physics work to suspend a load. But it can take quite a variable amount of chemical or electrical energy to create the couterbalancing force necessary to prevent the object from moving. A floor or a rope gets it for free; an electromagnet has to spend more electricity to balance a heavier load (at the same distance), a rocket must consume more propellant to hover a heavier load, etc. None of them are doing work, in the physics sense, but that doesn't mean they aren't expending energy.

Ok, granted the slider cars aren't doing work to suspend their load. But do they get the necessary suspensive force for 'free' like the ground, or rails, or a bridge does? Or do they have to constantly pay for it, like an electromagnet, a helicopter, or a rocket?
If the later how much do they have to pay relative to the capacity of their spell accumulators?
If the former how does the magic make thin air act like a (basically) frictionless solid surface?

That amount of energy is undefined, so the answer has to be no.

Since there isn't anything being lifted, that integral George referred to can't be evaluated.

Worse, since you've introduced 'for that length of time' you are no longer talking about energy, you're talking about power. Energy doesn't care about time, so it's the same for a short time as a long one.

However, to give you an idea of the "power" involved, let's assume that your scenario does have physical meaning and run the numbers. I'll use a pod towed by a super horse, to keep things smallish. Come to think of it, we should run it twice, just to illustrate something significant.

Mass = 5,000kg

Height = 2m

E=mgh= 5000x9.8x2 =98,000 Joules

For a one hour lift:

P=E/t = 98,000/3,600 = 27.222Watts

For ten days:

P = 98,000/864,000 = 113milliWatts

Notice how small these numbers are? More importantly, notice how it drops as the length of time goes up? Whatever it is you've defined, it doesn't seem to behave quite the way you seem to be expecting.

Those 98,000 Joules, BTW? That's the energy needed to boil a largish cup of water.

PeterZ wrote:I understand now. Thank you.

I still have to ask how much energy is used to prevent a body not supported by anything physical from accelerating at 32 feet/sec/sec? Assuming gravity is not altered in any way and energy/force is required to do the levitating.

Wouldn't that be the amount of energy required to perform the work of lifting that body at a rate of 32 feet/sec/sec for however long the spell is in effect for? After all, absent the supporting energy the body would accelerate at 32 feet/sec^2. The default state here is acceleration and to counter that state would require and equal and opposite force. As I recall Newton's First Law of Motion pretty much covers this.

All this goes out the window if arcane energy is used to change gravity so that it no longer pulls on the body.

phillies wrote:Your definition of work might be magically correct, but is not consistent with the definition of work used in our physics. The work done by my floor to hold me up is, well, zero, and the floor of my den does not get tired or run out of energy holding me up.

Work = \integral_start^Finish F(x) dx (for 1-Dimensional motion)

For levitation the displacement parallel to F is zero, so the work is zero.

George
D.Sc., MIT (Physics) '73.

98,000 Joules are required to lift the 5,000 kg mega-horse 2 meters. Absent more energy the mega horse will fall to the ground in less than 1 second, about .4 seconds.

So the power to maintain that altitude for 1 second is 245,000 Watts. To maintain that altitude for 1 hour is 245,000 x 3600 = 8.82x 10^8 Watts. For 10 days is 2.1168 x 10^11 Watts.

That is the logic I believe applies. I may be wrong, but that would mean gravity would cease to work for the duration of the levitation period.

Louis R wrote:That amount of energy is undefined, so the answer has to be no.

Since there isn't anything being lifted, that integral George referred to can't be evaluated.

Worse, since you've introduced 'for that length of time' you are no longer talking about energy, you're talking about power. Energy doesn't care about time, so it's the same for a short time as a long one.

However, to give you an idea of the "power" involved, let's assume that your scenario does have physical meaning and run the numbers. I'll use a pod towed by a super horse, to keep things smallish. Come to think of it, we should run it twice, just to illustrate something significant.

Mass = 5,000kg

Height = 2m

E=mgh= 5000x9.8x2 =98,000 Joules

For a one hour lift:

P=E/t = 98,000/3,600 = 27.222Watts

For ten days:

P = 98,000/864,000 = 113milliWatts

Notice how small these numbers are? More importantly, notice how it drops as the length of time goes up? Whatever it is you've defined, it doesn't seem to behave quite the way you seem to be expecting.

Those 98,000 Joules, BTW? That's the energy needed to boil a largish cup of water.

The problem I have with this calculation (and Peter's) is that you're using the potential energy of height. A helicopter (to use the previous example) doesn't care about height. it can be 100 or 1,000 or 10,000 feet off the ground, and it takes just as much additional force from the rotor (and thus, additional energy/fuel) to hold it up. Textev is that levitation accumulators can be used to let a dragon carry tons more, so they aren't "pushing" off the ground and the height probably has little or no relevance. So I don't believe any calculation that relies on height or time to reach the ground can be applicable. That's what I find so difficult about calculating this sort of thing.

I cannot believe that it can hover there for free, whatever the physics involved, and the textev specifically states that it can't. It also states that the higher the weight, the more the cost and/or the lower the time it can float is. Calculating how much magical energy it takes will depend on how much magical energy it takes to create a force that can counterbalance the weight, or to change the gravity of that much mass to zero. Unfortunately, I don't think we have enough information to make that calculation.

Louis R wrote:That amount of energy is undefined, so the answer has to be no.

Since there isn't anything being lifted, that integral George referred to can't be evaluated.

Worse, since you've introduced 'for that length of time' you are no longer talking about energy, you're talking about power. Energy doesn't care about time, so it's the same for a short time as a long one.

However, to give you an idea of the "power" involved, let's assume that your scenario does have physical meaning and run the numbers. I'll use a pod towed by a super horse, to keep things smallish. Come to think of it, we should run it twice, just to illustrate something significant.

Mass = 5,000kg

Height = 2m

E=mgh= 5000x9.8x2 =98,000 Joules

For a one hour lift:

P=E/t = 98,000/3,600 = 27.222Watts

For ten days:

P = 98,000/864,000 = 113milliWatts

Notice how small these numbers are? More importantly, notice how it drops as the length of time goes up? Whatever it is you've defined, it doesn't seem to behave quite the way you seem to be expecting.

Those 98,000 Joules, BTW? That's the energy needed to boil a largish cup of water.

PeterZ wrote:I understand now. Thank you.

I still have to ask how much energy is used to prevent a body not supported by anything physical from accelerating at 32 feet/sec/sec? Assuming gravity is not altered in any way and energy/force is required to do the levitating.

Wouldn't that be the amount of energy required to perform the work of lifting that body at a rate of 32 feet/sec/sec for however long the spell is in effect for? After all, absent the supporting energy the body would accelerate at 32 feet/sec^2. The default state here is acceleration and to counter that state would require and equal and opposite force. As I recall Newton's First Law of Motion pretty much covers this.

All this goes out the window if arcane energy is used to change gravity so that it no longer pulls on the body.

You are indeed completely wrong.

Essentially, you are confounding energy with force - they are not synonyms. I have no idea where you get the idea that no force can be applied without expending energy, but that is plainly and manifestly incorrect. In fact, work is done if, and only if, an object moves under the influence of a force. Otherwise zilch, zip, nada, rien! I present objects in orbit around a central mass as proof - in Newton's formulation, that is the classic example of action-at-a-distance. There are many, many others out there. [and yes, I am entirely aware that in Einstein's formulation there is no such thing as action at a distance and something entirely different is happening. So what? Switch to one of the other formulations and even weirder stuff starts popping out of the woodwork, and in any case it can be demonstrated that under the conditions we're concerned with, they are all formally equivalent - which is a problem, actually, but not one that need trouble us. All of them tell us that no work has been done.]

And, actually, gravity clearly doesn't cease to work during levitation. The results of that would be quite different, probably spectacular, and very likely awkward for all concerned. It is probably far more correct to state that the levitation spell establishes a meta-physical connection between object and the ground below it such that the reaction of the ground to the weight of the object continues to act on it in the absence of physical contact. IOW, a second action-at-a-distance has been added so that the object is in static equilibrium somewhere other than on the ground. Any magical power - which needn't have even a conceptual relationship with the physical quantity defined as power - expended maintaining that condition is dissipated in the internal workings of the spell. I can think of several ways that might occur; there's no way to quantify any of them.

PS: I did warn you that that calculation was physically meaningless. I made the mistake of waving some numbers at you, and you've buggered everything up again: the product of power and time is energy, not more power. The product of energy and time is not a physical quantity.

PeterZ wrote:98,000 Joules are required to lift the 5,000 kg mega-horse 2 meters. Absent more energy the mega horse will fall to the ground in less than 1 second, about .4 seconds.

So the power to maintain that altitude for 1 second is 245,000 Watts. To maintain that altitude for 1 hour is 245,000 x 3600 = 8.82x 10^8 Watts. For 10 days is 2.1168 x 10^11 Watts.

That is the logic I believe applies. I may be wrong, but that would mean gravity would cease to work for the duration of the levitation period.

Louis R wrote:That amount of energy is undefined, so the answer has to be no.

Since there isn't anything being lifted, that integral George referred to can't be evaluated.

Worse, since you've introduced 'for that length of time' you are no longer talking about energy, you're talking about power. Energy doesn't care about time, so it's the same for a short time as a long one.

However, to give you an idea of the "power" involved, let's assume that your scenario does have physical meaning and run the numbers. I'll use a pod towed by a super horse, to keep things smallish. Come to think of it, we should run it twice, just to illustrate something significant.

Mass = 5,000kg

Height = 2m

E=mgh= 5000x9.8x2 =98,000 Joules

For a one hour lift:

P=E/t = 98,000/3,600 = 27.222Watts

For ten days:

P = 98,000/864,000 = 113milliWatts

Notice how small these numbers are? More importantly, notice how it drops as the length of time goes up? Whatever it is you've defined, it doesn't seem to behave quite the way you seem to be expecting.

Those 98,000 Joules, BTW? That's the energy needed to boil a largish cup of water.

I would like to complicate the issue. Text describes 2 different effects from levitation spells. The levitating chairs and tables of the Fallen Timbers negotiations did not function as if they and the objects or people on them had zero mass. Those tables and chairs were suspended by some force making them stable enough to sit on and rest items on.

The levitating spells used with transport dragons do appear to alter the impact of gravity on the mass within its area of effect.

Neither affect changes our inability to calculate the amount of energy involved or know the level of Gift required create and then charge the different spells.

That's why "power" is in quotes - the whole exercise is physically meaningless. I engaged in it simply to show that the original suggestion lead to a result that doesn't make any sense - that the longer you stay up, the less power you need to do it.

BTW, that helicopter is a perfect example of what I've been trying to get at; thank you.The energy is not being spent 'holding it up', it is spent moving the rotors against the drag force of the air they're moving in: it is being dissipated internally in the system that generates the force holding the aircraft in equilibrium. And in fact it does matter how high it is, although usually not enough to notice - drag is a function of air density and temperature, and decreases with height at a different rate than the lift generated by the blades.

bkwormlisa wrote:The problem I have with this calculation (and Peter's) is that you're using the potential energy of height. A helicopter (to use the previous example) doesn't care about height. it can be 100 or 1,000 or 10,000 feet off the ground, and it takes just as much additional force from the rotor (and thus, additional energy/fuel) to hold it up. Textev is that levitation accumulators can be used to let a dragon carry tons more, so they aren't "pushing" off the ground and the height probably has little or no relevance. So I don't believe any calculation that relies on height or time to reach the ground can be applicable. That's what I find so difficult about calculating this sort of thing.

I cannot believe that it can hover there for free, whatever the physics involved, and the textev specifically states that it can't. It also states that the higher the weight, the more the cost and/or the lower the time it can float is. Calculating how much magical energy it takes will depend on how much magical energy it takes to create a force that can counterbalance the weight, or to change the gravity of that much mass to zero. Unfortunately, I don't think we have enough information to make that calculation.