Making Levitation AWESOME., Trade weight for speed, and vice versa. |
Making Levitation AWESOME., Trade weight for speed, and vice versa. |
Jun 4 2006, 07:38 AM
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#1
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Dragon Group: Members Posts: 4,589 Joined: 28-November 05 Member No.: 8,019 |
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. |
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Jun 4 2006, 07:42 AM
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#2
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Great Dragon Group: Members Posts: 5,430 Joined: 10-January 05 From: Fort Worth, Texas Member No.: 6,957 |
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.
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Jun 4 2006, 07:43 AM
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#3
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Dragon Group: Members Posts: 4,589 Joined: 28-November 05 Member No.: 8,019 |
OOOOOOOHHHHH yes. Magic kicks so much ass.
Maybe you could have Greater Levitation, where you can do this but for +2 drain. |
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Jun 4 2006, 08:24 AM
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#4
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Great Dragon Group: Members Posts: 5,889 Joined: 3-August 03 From: A CPI rank 1 country Member No.: 5,222 |
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.
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Jun 4 2006, 08:26 AM
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#5
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Dragon Group: Members Posts: 4,589 Joined: 28-November 05 Member No.: 8,019 |
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. |
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Jun 4 2006, 08:27 AM
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#6
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Runner Group: Members Posts: 2,598 Joined: 15-March 03 From: Hong Kong Member No.: 4,253 |
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. |
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Jun 4 2006, 08:28 AM
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#7
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Dragon Group: Members Posts: 4,589 Joined: 28-November 05 Member No.: 8,019 |
Not... awesome... enough!!! :grr:
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Jun 4 2006, 08:42 AM
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#8
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Dragon Group: Members Posts: 4,664 Joined: 21-September 04 From: Arvada, CO Member No.: 6,686 |
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.
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Jun 4 2006, 08:47 AM
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#9
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Runner Group: Members Posts: 2,598 Joined: 15-March 03 From: Hong Kong Member No.: 4,253 |
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. |
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Jun 4 2006, 04:58 PM
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#10
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Moving Target Group: Members Posts: 316 Joined: 18-April 05 From: France Member No.: 7,343 |
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. |
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Jun 4 2006, 06:27 PM
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#11
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Moving Target Group: Members Posts: 626 Joined: 1-March 04 Member No.: 6,112 |
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. |
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Jun 5 2006, 12:51 PM
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#12
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Moving Target Group: Members Posts: 158 Joined: 10-April 06 Member No.: 8,448 |
Erased.
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Jun 5 2006, 12:56 PM
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#13
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Target Group: Members Posts: 24 Joined: 26-May 06 Member No.: 8,606 |
"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? |
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Jun 5 2006, 01:23 PM
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#14
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Neophyte Runner Group: Members Posts: 2,073 Joined: 23-August 04 Member No.: 6,587 |
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) |
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Jun 5 2006, 01:26 PM
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#15
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Incertum est quo loco te mors expectet; Group: Dumpshocked Posts: 6,546 Joined: 24-October 03 From: DeeCee, U.S. Member No.: 5,760 |
Speaking for myself, one movement power per customer. No shoes, no shirt, no service.
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Jun 5 2006, 01:31 PM
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#16
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Moving Target Group: Members Posts: 158 Joined: 10-April 06 Member No.: 8,448 |
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. |
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Jun 5 2006, 02:43 PM
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#17
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Runner Group: Members Posts: 2,598 Joined: 15-March 03 From: Hong Kong Member No.: 4,253 |
Kintetic energy = (1/2) * m * v^2 for 'newtonian' objects |
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Jun 5 2006, 03:22 PM
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#18
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Great Dragon Group: Members Posts: 7,089 Joined: 4-October 05 Member No.: 7,813 |
are you perhaps thinking of momentum? because kinetic energy is indeed k=1/2mV^2 |
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Jun 5 2006, 03:59 PM
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#19
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Neophyte Runner Group: Members Posts: 2,073 Joined: 23-August 04 Member No.: 6,587 |
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 |
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Jun 5 2006, 04:15 PM
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#20
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Great Dragon Group: Members Posts: 7,089 Joined: 4-October 05 Member No.: 7,813 |
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. |
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Jun 5 2006, 04:32 PM
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#21
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Runner Group: Members Posts: 2,598 Joined: 15-March 03 From: Hong Kong Member No.: 4,253 |
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. |
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Jun 5 2006, 04:52 PM
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#22
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Neophyte Runner Group: Members Posts: 2,073 Joined: 23-August 04 Member No.: 6,587 |
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 |
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Jun 5 2006, 05:49 PM
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#23
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Great Dragon Group: Members Posts: 5,889 Joined: 3-August 03 From: A CPI rank 1 country Member No.: 5,222 |
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. |
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Jun 5 2006, 09:17 PM
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#24
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Moving Target Group: Members Posts: 834 Joined: 30-June 03 Member No.: 4,832 |
Scaling the levitate the otherway, you could lift an arcology up very very slowly, assuming you bypass it's OR threshold.
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Jun 6 2006, 09:49 PM
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#25
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Runner Group: Members Posts: 2,556 Joined: 26-February 02 From: Seattle Member No.: 98 |
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. |
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