emo samurai
Jun 4 2006, 07:38 AM
Force 1 levitation lets you move 250 KG at 1 meter per round. If I were lifting anything lighter than a car, I would gladly trade weight for speed; instead of 250 KG at 1 meter per round, I would have it move 125 KG at 2 meters per round, or something like that. Would you allow this?
I imagine that at the higher forces, you'd be flying around like DBZ characters; I hope you know enough about me by now to realize that I don't care.
James McMurray
Jun 4 2006, 07:42 AM
I wouldn't allow it, but it would probably fit right in with your game. would you let it stack with the movement power? I don't have a book handy but I imagine you could get some scary movement rates with a force 8-ish spirit and spell.
emo samurai
Jun 4 2006, 07:43 AM
OOOOOOOHHHHH yes. Magic kicks so much ass.
Maybe you could have Greater Levitation, where you can do this but for +2 drain.
Austere Emancipator
Jun 4 2006, 08:24 AM
Be sure to let it scale all the way down. 250kg @ 1m/CT -> 0.015kg steel chunk at 5464m/s = 235.5 grains at 17926fps = kinda like a sporting rifle, only at over 6 times the velocity/36 times the kinetic energy. Should be enough to penetrate at least 4" of armor steel, or, with a correctly shaped steel chunk, to spectacularly fuck up any metahuman.
emo samurai
Jun 4 2006, 08:26 AM
OHHHHHH yes. That would rock so hard. And it wouldn't arc, as long as it stayed in your LOS.
You did not in any way put a monkey wrench in the power trip with that post.
Crusher Bob
Jun 4 2006, 08:27 AM
It is still possible to achieve pretty good speeds with SR4 levitate, for example a mage with a magic of 6, a force 4 power focus, and rolling 15 dice (and getting 5 hits) goes at 60 km/h. Adding his edge (of 3) and rolling 18 exploding dice moves him at 84 km/h.
Sure it's not the high subsonic speeds that previous editions mages could achieve, but its still pretty respectable.
emo samurai
Jun 4 2006, 08:28 AM
Not... awesome... enough!!!
Tarantula
Jun 4 2006, 08:42 AM
Maybe, you could have a levitate brain spell, and just levitate their brains right out of their heads! 3 guesses to who gave me the inspiration for that idea.
Crusher Bob
Jun 4 2006, 08:47 AM
Ok, so figure out how fast you want to go
chump mage: magic 4, 8 dice, using edge 11 dice
Starting mage: magic 5, 10 dice, using edge 13 dice
Starting mage w/ p focus 3, magic 8, 13 dice, using edge 16 dice
Powerful mage magic 8, 12 dice, using edge 16 dice
Powermage + focus 4, magic 12, 16 dice, using edge 20 dice
Top'o the heap magic 10, 14 dice, using edge 20 dice
+ focus 6 magic 16, 20 dice, using edge 26 dice.
so the scale looks something like this:
chumps (4 x 4) levitate factor 16
top 'o the heap (w/focus w/o edge) (16x7) levitate factor 112
assuming we want our top o' the heap mage to top out around the speed of sound (~1200 km/h at sea level), we'll fudge and say around 10 km/h x the 'letivate factor).
This would make a chump using the same spell go around 160 km/h.
For reference, the SR4 stock speed multiple is 1.2 km/h per levitate factor. This means that the 'supa fly' spells would have to work ~ 8.33 times as well as regular levitate.
Taki
Jun 4 2006, 04:58 PM
QUOTE (emo samurai) |
Force 1 levitation lets you move 250 KG at 1 meter per round. If I were lifting anything lighter than a car, I would gladly trade weight for speed; instead of 250 KG at 1 meter per round, I would have it move 125 KG at 2 meters per round, or something like that. Would you allow this ? |
Actually I would'nt allow to do it this way :
Because the ammount of destruction, and the real energy is represented by cinetic energy, which is equal to 1/2 * mass * movement^2
Thus, by divided simply mass by 2 and multiplying movement by 2 you actually multiply the puissance by 2.
I think it woul be more balanced if you multiply speed by two and divide the mass by 4.
Toptomcat
Jun 4 2006, 06:27 PM
That'd permit relatively low-Force Levitation to accelerate a sufficiently small grain of dust to ≈ c.
No. Bad emo samurai. No soup for you.
UndeadPoet
Jun 5 2006, 12:51 PM
Erased.
irinoxx
Jun 5 2006, 12:56 PM
QUOTE (UndeadPoet) |
QUOTE (emo samurai @ Jun 4 2006, 02:38 AM) | I imagine that at the higher forces, you'd be flying around like DBZ characters; I hope you know enough about me by now to realize that I don't care. |
Use a decent force levitation, use edge on the cast, let a mighty bound spirit of men sustain his movement-power on you, et voila. Or how that's written correctly. I get confused with the many ' and ` in french.
Well, this way, you can easily get faster than any vehicle in the game, except this plane with 1000 metres/round. And, yeah, that is like DBZ. God, I hate DBZ!
|
"voilà".
By the way, how do you handle multiple spirits using their Movement power on a character?
- multiply to craaaaazy speeds?
- add their Force, then multiply?
- take biggest spirit Force, add 1 for every helping spirit, then multiply?
- can't have more than one spirit helping movement?
Edward
Jun 5 2006, 01:23 PM
QUOTE (Taki) |
Because the ammount of destruction, and the real energy is represented by cinetic energy, which is equal to 1/2 * mass * movement^2
|
that is not a correct formula for kinetic energy.
Kinetic energy = mass * speed
When you start considering relativistic speeds (speeds that constitute a significant fraction of the speed of light) the square of the speed may become significant but I don’t think we should allow anything the PCs interact with in shadow run to move at relativistic speeds (no object in the solar system moves at a relativistic speed, with the exception of some forms of high energy radiation and light)
nezumi
Jun 5 2006, 01:26 PM
Speaking for myself, one movement power per customer. No shoes, no shirt, no service.
UndeadPoet
Jun 5 2006, 01:31 PM
QUOTE (nezumi) |
Speaking for myself, one movement power per customer. No shoes, no shirt, no service. |
Yeah, right. I mean, nothing in SR4 actually stacks. Of course, reflexbooster and enhanced reaction do stack, but not 2 reflexboosters or 2 enhanced reactions(correct plural? ^^).
So, ironixx, your last option.
Crusher Bob
Jun 5 2006, 02:43 PM
QUOTE (Edward) |
Kinetic energy = mass * speed
|
Kintetic energy = (1/2) * m * v^2 for 'newtonian' objects
Jaid
Jun 5 2006, 03:22 PM
QUOTE (Edward) |
QUOTE (Taki) | Because the ammount of destruction, and the real energy is represented by cinetic energy, which is equal to 1/2 * mass * movement^2
|
that is not a correct formula for kinetic energy.
Kinetic energy = mass * speed
When you start considering relativistic speeds (speeds that constitute a significant fraction of the speed of light) the square of the speed may become significant but I don’t think we should allow anything the PCs interact with in shadow run to move at relativistic speeds (no object in the solar system moves at a relativistic speed, with the exception of some forms of high energy radiation and light)
|
are you perhaps thinking of momentum?
because kinetic energy is indeed k=1/2mV^2
Edward
Jun 5 2006, 03:59 PM
Ok, a quick web search shows your right. It doesn’t make any sense to me but your right.
A model rocket engine has a fixed amount on chemical energy witch will translate into a fixed amount of kinetic energy
That would mean it takes more fuel (chemical energy) to accelerate from 100kphto 200kph than from 0 to 100kph (3 times as much fuel). In the absence of wind resistance this doesn’t make sense, to start with what do you use as your reference point to determine your speed,
Edward
Jaid
Jun 5 2006, 04:15 PM
if you don't mind my asking, why precisely does that not make sense?
i would certainly say that velocity increases (as compared to energy expenditure) follow some kind of diminishing returns type equation. certainly, if i run at half my maximum speed, i can do so for more than twice as long as i can run at 100% of my maximum speed, IMO.
what i find fascinating is that it works out to such a simple equation... the constant is only 1/2, not something wacky like pi, and the exponent is just 2, not something wacky like the square root of 5.
Crusher Bob
Jun 5 2006, 04:32 PM
QUOTE (Jaid @ Jun 6 2006, 12:15 AM) |
what i find fascinating is that it works out to such a simple equation... the constant is only 1/2, not something wacky like pi, and the exponent is just 2, not something wacky like the square root of 5. |
Take a look at the units of the equasion:
(energy is expressed in joules with is defined as one newton * meter)
A newton (a measure of thrust) is kg * m /s^2
so a joule is kg * m^2/s^2
on the other side we have:
(1/2) kg (m/s)^2
Taking the derivitave of (1/2) kg m^2/s^2 gives us kg m/s^2
giving us another of Newton's equasions force = mass * acceleration.
So basically, all of Newton's (the man, not the unit of measurement) equasions
Ke = (1/2)mv^2
d = (1/2)at^2
F=ma
etc
are simply the same thing but with some calculus applied.
Edward
Jun 5 2006, 04:52 PM
First the multiplier is simple because the units are all metric, with metric the constants always come out nicely as long as you don’t actually include a natural physical phenomena (such as the charge of an electron or the speed of light) metric defined one unit (the meter) and all other units where created based on that, this includes the measure of speed, the measure of mass and the measure of energy
Running is a bad example when you run you have to change the velocity of your legs more often the faster you move. Its /almost/ a fixed amount of energy per step (there are also problems getting energy to the muscles and dealing with waste chemicals).
Consider 3 objects in the absence of an atmosphere or any other resistance
All 3 objects(A B C) have mass 1 and are at the arbitrary stationary
2 objects(A B) are affected buy a force of one unit for one unit of time. They gain one unit of speed and thus one unit of kinetic energy.
One of those 2 objects (A) is again affected buy a force, the same force in the same direction for the same time. When observed from object B object A now has speed one and thus kinetic energy one
However if the same force on A was observed from C A was already moving so it gains one unit of energy for a total of 2., reversing the formula we get a speed of the square root of 2
Thus speed a-b =1
Speed b-c = 1
Speed a-c = root 2
1+1=root 2
Now strange stuff like this is supposed to happen under the theory of relativity but not until you move at an appreciable fraction of the speed of light.
That is why it doesn’t make sense
Edward
Austere Emancipator
Jun 5 2006, 05:49 PM
QUOTE (Edward) |
Consider 3 objects in the absence of an atmosphere or any other resistance
All 3 objects(A B C) have mass 1 and are at the arbitrary stationary
2 objects(A B) are affected buy a force of one unit for one unit of time. They gain one unit of speed and thus one unit of kinetic energy.
One of those 2 objects (A) is again affected buy a force, the same force in the same direction for the same time. When observed from object B object A now has speed one and thus kinetic energy one
However if the same force on A was observed from C A was already moving so it gains one unit of energy for a total of 2., reversing the formula we get a speed of the square root of 2 |
m = mass
d = distance
t = time
Looking at Object A from Object C, after the first application of The Force, A had a velocity of 1d/t, and kinetic energy equal to 0.5 x 1m x (1d/t)^2 = 0.5md^2/t^2. After the second application of The Force, Object A has a velocity (relative to Object C) of 2d/t and kinetic energy equal to 0.5 x 1m x (2d/t)^2 = 2md^2/t^2. Thus, with the second application of force, Object A gained 1.5 units of kinetic energy.
I have no idea what you're trying to say with the rest of that. But if you've really got a problem with this, I suggest you take it to a high school physics teacher and not an internet forum for a RPG.
Lagomorph
Jun 5 2006, 09:17 PM
Scaling the levitate the otherway, you could lift an arcology up very very slowly, assuming you bypass it's OR threshold.
Shrike30
Jun 6 2006, 09:49 PM
You've got an object sitting in a neutral environment (no air, no gravity, nothing to screw with this example). It weighs 1 kg, it's doing 0 mps. 0 momentum, 0 KE. You've got another object in that same environment. It weighs 1 kg, it's doing 10 mps. Momentum is 10 (momentum = mass * velocity, 1x10=10), KE is 50 (KE = .5 M * V^2, .5x1x(10x10) = .5(100) = 50).
These two objects collide efficiently, and become a single 2 kg object travelling at 5 mps. Your momentum is still 10 (2x5 = 10), as momentum is conserved. However, your KE has just plummeted to 25 (.5x2x(5x5) = 1x(25) = 25). Energy is conserved as well, so where did that other 25 joules go? Heat. Your objects are now 25 joules warmer.
When you've got a rocket firing in space, it sends reaction mass in one direction at a given velocity, the rocket in the other. The amount of velocity gained by the rocket is directly linked to the momentum of the reaction mass going outwards, not to the kinetic energy of the reaction mass.
Edward
Jun 11 2006, 03:58 PM
QUOTE (Austere Emancipator) |
after the first application of The Force, A had a velocity of 1d/t, and kinetic energy equal to 0.5 x 1m x (1d/t)^2 = 0.5md^2/t^2.
with the second application of force, Object A gained 1.5 units of kinetic energy. |
Much sniping above
How can 2 applications of the same force for the same time result in different changes to energy.
That is my problem.
As to lifting the arcology. It only has an OR of 3 or 4. Any competent magician will be able to lift it.
Edward
Austere Emancipator
Jun 11 2006, 04:24 PM
The same force, different applications.
Let's assume The Force is/has a force of 1md/t^2. When it accelerates Object A, with mass 1m, to a velocity of 1d/t, it does 0.5md^2/t^2 of work -- The Force is applied on Object A for a distance of 0.5d -- increasing Object A's kinetic energy from 0md^2/t^2 to 0.5md^2/t^2. When it accelerates Object A from 1d/t to 2d/t, it does 1.5md^2/t^2 of work, applying 1md/t^2 of force on Object A for a distance of 1.5d, increasing the kinetic energy of Object A from 0.5md^2/t^2 to 2md^2/t^2.
And, again, it would be better to resolve these issues of yours with a physics teacher than someone who almost flunked high school physics on an RPG forum.
Shrike30
Jun 12 2006, 06:23 PM
Two applications of the same amount of force to objects with different masses will result in a matching change in
momentum, not energy. Energy's relation to mass is geometric, but it's relation to velocity is exponential (as I laid out about 3 posts ago). If you don't buy this, I suggest taking some physics courses or looking it up online.
Wikipedia has a decent article on this.
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