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ArchAngle
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Posted - 2004.07.08 07:55:00 -
[1]
let me ask you this lets say a jet movign at 1000 mph is going foward and a steel boling ball moving strait towards it and hits it in the nose what will happen to the jet. This is kinda complex but it stats what will happen in a this exaample and also what will happen to a missle when hitting a object.
PART ONE
One of the common arguments given for kinetic energy being ¢mv¦ is that it shows a net increase when energy is added to a system, while mv does not. The model for this point is an explosion blasting two objects in opposite directions. The question is whether there is a net increase in momentum.
The quantity mv increases by the same amount in two opposite directions, because for every force there must be an equal and opposite force. The increase in one direction is given a minus sign; and when added to the mv in the other direction, the total is zero.
The quantity ¢mv¦, however, shows a net increase. Its velocity is squared, which converts any negative quantity to a positive quantity; and adding two positives always shows a net increase.
That argument is a fallacy. It is not valid to put a minus sign by momentum when quantitating energy addition, because there is no such thing as a negative quantity of momentum. (In determining velocities, the negative sign has a different purpose.) An increase of momentum in two opposite directions is an increase in momentum. Therefore, momentum should be quantitated in absolute values when relating to energy.
Here's another way of stating it. The original question is total quantity of energy. The minus sign changes the question to vectorial quality. After adding, it is then reinterpreted in terms of total quantity. Switching back and forth between total quantity and vectorial quality is not valid.
The assumption that energy must show a directional increase relative to an external reference frame is a fallacy. This point is demonstrated when heating a piece of metal. There is no directional increase in energy.
Energy exists regardless of the direction of movement. Heat demonstrates this point. It is a randomization of motion. And it is called energy. To say that negative momentum cancels positive momentum is the same as saying half of the heat cancels the other half of the heat.
For a correct analysis, energy addition must be evaluated relative to the point where the forces act or the impact point. Momentum increases in both directions relative to the impact point, when energy is added.
In analyzing collisions, there is often no negative velocity, because the center of mass may be moving at a high velocity relative to an external reference frame resulting in both of the equal and opposite momentums having a positive velocity. Regardless of whether there is a negative velocity, no net momentum change occurs relative to an external reference frame, when energy is added, even though the momentums do change relative to the center of mass.
There is a contradiction in saying that the net energy increases, while the net momentum does not. Momentum is the motion of a mass. If the motion of the mass has no net increase, how could there be a net change in the kinetic energy, which is said to be the energy of motion?
As a matter of fact, kinetic energy as ¢mv¦ is not in the motion of the mass, because no mass can move at velocity squared. Its motion is first said to be its velocity (Momentum and velocity are both motion.); and its velocity is not its velocity squared. So ¢mv¦ is an abstraction apart from the motion of the mass. The question is, can energy really be an abstraction, if it is used in discrete quantities as fuel? Fuel is more than an abstraction; so the energy defined by the equation must be more than an abstraction. Momentum is perceivable as mass and velocity; but ¢mv¦ is not.
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ArchAngle
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Posted - 2004.07.08 07:55:00 -
[2]
let me ask you this lets say a jet movign at 1000 mph is going foward and a steel boling ball moving strait towards it and hits it in the nose what will happen to the jet. This is kinda complex but it stats what will happen in a this exaample and also what will happen to a missle when hitting a object.
PART ONE
One of the common arguments given for kinetic energy being ¢mv¦ is that it shows a net increase when energy is added to a system, while mv does not. The model for this point is an explosion blasting two objects in opposite directions. The question is whether there is a net increase in momentum.
The quantity mv increases by the same amount in two opposite directions, because for every force there must be an equal and opposite force. The increase in one direction is given a minus sign; and when added to the mv in the other direction, the total is zero.
The quantity ¢mv¦, however, shows a net increase. Its velocity is squared, which converts any negative quantity to a positive quantity; and adding two positives always shows a net increase.
That argument is a fallacy. It is not valid to put a minus sign by momentum when quantitating energy addition, because there is no such thing as a negative quantity of momentum. (In determining velocities, the negative sign has a different purpose.) An increase of momentum in two opposite directions is an increase in momentum. Therefore, momentum should be quantitated in absolute values when relating to energy.
Here's another way of stating it. The original question is total quantity of energy. The minus sign changes the question to vectorial quality. After adding, it is then reinterpreted in terms of total quantity. Switching back and forth between total quantity and vectorial quality is not valid.
The assumption that energy must show a directional increase relative to an external reference frame is a fallacy. This point is demonstrated when heating a piece of metal. There is no directional increase in energy.
Energy exists regardless of the direction of movement. Heat demonstrates this point. It is a randomization of motion. And it is called energy. To say that negative momentum cancels positive momentum is the same as saying half of the heat cancels the other half of the heat.
For a correct analysis, energy addition must be evaluated relative to the point where the forces act or the impact point. Momentum increases in both directions relative to the impact point, when energy is added.
In analyzing collisions, there is often no negative velocity, because the center of mass may be moving at a high velocity relative to an external reference frame resulting in both of the equal and opposite momentums having a positive velocity. Regardless of whether there is a negative velocity, no net momentum change occurs relative to an external reference frame, when energy is added, even though the momentums do change relative to the center of mass.
There is a contradiction in saying that the net energy increases, while the net momentum does not. Momentum is the motion of a mass. If the motion of the mass has no net increase, how could there be a net change in the kinetic energy, which is said to be the energy of motion?
As a matter of fact, kinetic energy as ¢mv¦ is not in the motion of the mass, because no mass can move at velocity squared. Its motion is first said to be its velocity (Momentum and velocity are both motion.); and its velocity is not its velocity squared. So ¢mv¦ is an abstraction apart from the motion of the mass. The question is, can energy really be an abstraction, if it is used in discrete quantities as fuel? Fuel is more than an abstraction; so the energy defined by the equation must be more than an abstraction. Momentum is perceivable as mass and velocity; but ¢mv¦ is not.
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ArchAngle
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Posted - 2004.07.08 08:14:00 -
[3]
*PART TWO*
Simple Collisions.
To demonstrate the principles of collision analysis, a simple type of collision will be analyzed, which is a head-on elastic collision of two masses. The predictability of the collision is dependent upon the fact that the center of mass (CM) maintains the same velocity after the collision as before. Also, the momentum relative to the center of mass stays the same for each mass but reverses directions.
Mass 1 is 8kg moving right at 50m/s. Mass 2 is 4kg moving left at 10m/s. The net momentum is the sum of the two momentums, when the velocity left is given a negative sign.
NM = m1v1 + m2v2 = 8(50) + 4(-10) = 360
The velocity of the center of mass (vCM) equals the total momentum divided by the total mass.
vCM = 360/12 = 30
The initial velocity of each mass relative to CM is its original velocity minus vCM.
m1: 50 - 30 = 20 m2: -10 -30 = -40
Changing the sign produces the same velocity relative to CM after the collision. The velocity is then converted to the original reference frame by adding the velocity of CM.
m1: -20 + 30 = 10 m2: 40 + 30 = 70
A simpler procedure is possible by analyzing the velocities only. The final velocity of each mass will be twice the velocity of CM minus its velocity before the collision.
2vCM - v(before) = v(after)
m1: 2(30) - 50 = 10 m2: 2(30) - (-10) = 70
With these numbers, the mv and ¢mv¦ are calculated.
______________________________________
] m1 ] m2 ] total ]
]___________]___________]_____________]
] mv ]¢mv¦ ] mv ]¢mv¦ ] mv ] ¢mv¦ ]
]___________]___________]_____________]
]400 ]10,000] -40 ] 200 ] 360 ] 10,200] <--before
]-------------------------------------]
]80 ] 400 ] 280 ]9,800] 360 ] 10,200] <--After
---------------------------------------
This example only shows that mv and ¢mv¦ are both conserved during elastic collisions (the total columns). The bigger question is what happens when energy is added to the system.
Adding Energy.
To demonstrate the addition of energy, the following example will start with the 8kg and 4kg masses combined and moving right at 30m/s; and then an explosion will separate them with a force averaging 1600N (newtons) for 0.1 seconds. First acceleration will be determined as a = F/m; and then velocity will be determined as v = at. (Force left is minus.)
m1: a = -1600/8 = -200 m2: a = 1600/4 = 400
m1: v = -200(0.1) = -20 m2: v = 400(0.1) = 40
These velocities are relative to CM; so they must be added to vCM to get velocities relative to an external reference frame.
m1: -20 + 30 = 10 m2: 40 + 30 = 70
So the result is this:
The resulting levels of mv and ¢mv¦ are:
______________________________________
] m1 ] m2 ] total ]
]___________]___________]_____________]
] mv ]¢mv¦ ] mv ]¢mv¦ ] mv ] ¢mv¦ ]
]___________]___________]_____________]
]240 ] 3,600] 120 ]1,800] 360 ] 5,400 ] <--before
]-------------------------------------]
]80 ] 400 ] 280 ]9,800] 360 ] 10,200] <--After
---------------------------------------
The table shows that relative to an external reference frame the energy added by the explosion increased the amount of ¢mv¦ but not the mv. (360mv before and after). The usual assumption is that this demonstrates that adding energy to a system does not change momentum. However, relative to the impact point, the momentum went from 0 to 160 in each direction.
The Corrected Concept.
So the question is, must enegy be able to change relative to an external reference frame; or is it something that only changes relative to impact points? Forces can only exist relative to impact pointsùnot relative to an external reference frame, which is relative and infinitely variable. So the correct definition of kinetic energy should be the change in momentum relative to impact points or the points where the forces act.*
The collision analysis is not a major proof in itself; but the perspective on collisions Would be improved with the correct definition of energy. The same is true of most points being made here. The mathematical proof is in applying rockets to the falling object issue, which is shown in the rocket section.
--------------------------------------------------------------------------------
*Notice in the rocket section that the analysis of power by erroneous concepts is made relative to the point where the forces act by using the separation velocity of the exhaust as the reference. This strange twist allows the rocket equations to be balanced with the erroneous definitions. It also contradicts the premise that energy addition must occur relative to an external reference frame by the erroneous concepts.
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ArchAngle
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Posted - 2004.07.08 08:14:00 -
[4]
*PART TWO*
Simple Collisions.
To demonstrate the principles of collision analysis, a simple type of collision will be analyzed, which is a head-on elastic collision of two masses. The predictability of the collision is dependent upon the fact that the center of mass (CM) maintains the same velocity after the collision as before. Also, the momentum relative to the center of mass stays the same for each mass but reverses directions.
Mass 1 is 8kg moving right at 50m/s. Mass 2 is 4kg moving left at 10m/s. The net momentum is the sum of the two momentums, when the velocity left is given a negative sign.
NM = m1v1 + m2v2 = 8(50) + 4(-10) = 360
The velocity of the center of mass (vCM) equals the total momentum divided by the total mass.
vCM = 360/12 = 30
The initial velocity of each mass relative to CM is its original velocity minus vCM.
m1: 50 - 30 = 20 m2: -10 -30 = -40
Changing the sign produces the same velocity relative to CM after the collision. The velocity is then converted to the original reference frame by adding the velocity of CM.
m1: -20 + 30 = 10 m2: 40 + 30 = 70
A simpler procedure is possible by analyzing the velocities only. The final velocity of each mass will be twice the velocity of CM minus its velocity before the collision.
2vCM - v(before) = v(after)
m1: 2(30) - 50 = 10 m2: 2(30) - (-10) = 70
With these numbers, the mv and ¢mv¦ are calculated.
______________________________________
] m1 ] m2 ] total ]
]___________]___________]_____________]
] mv ]¢mv¦ ] mv ]¢mv¦ ] mv ] ¢mv¦ ]
]___________]___________]_____________]
]400 ]10,000] -40 ] 200 ] 360 ] 10,200] <--before
]-------------------------------------]
]80 ] 400 ] 280 ]9,800] 360 ] 10,200] <--After
---------------------------------------
This example only shows that mv and ¢mv¦ are both conserved during elastic collisions (the total columns). The bigger question is what happens when energy is added to the system.
Adding Energy.
To demonstrate the addition of energy, the following example will start with the 8kg and 4kg masses combined and moving right at 30m/s; and then an explosion will separate them with a force averaging 1600N (newtons) for 0.1 seconds. First acceleration will be determined as a = F/m; and then velocity will be determined as v = at. (Force left is minus.)
m1: a = -1600/8 = -200 m2: a = 1600/4 = 400
m1: v = -200(0.1) = -20 m2: v = 400(0.1) = 40
These velocities are relative to CM; so they must be added to vCM to get velocities relative to an external reference frame.
m1: -20 + 30 = 10 m2: 40 + 30 = 70
So the result is this:
The resulting levels of mv and ¢mv¦ are:
______________________________________
] m1 ] m2 ] total ]
]___________]___________]_____________]
] mv ]¢mv¦ ] mv ]¢mv¦ ] mv ] ¢mv¦ ]
]___________]___________]_____________]
]240 ] 3,600] 120 ]1,800] 360 ] 5,400 ] <--before
]-------------------------------------]
]80 ] 400 ] 280 ]9,800] 360 ] 10,200] <--After
---------------------------------------
The table shows that relative to an external reference frame the energy added by the explosion increased the amount of ¢mv¦ but not the mv. (360mv before and after). The usual assumption is that this demonstrates that adding energy to a system does not change momentum. However, relative to the impact point, the momentum went from 0 to 160 in each direction.
The Corrected Concept.
So the question is, must enegy be able to change relative to an external reference frame; or is it something that only changes relative to impact points? Forces can only exist relative to impact pointsùnot relative to an external reference frame, which is relative and infinitely variable. So the correct definition of kinetic energy should be the change in momentum relative to impact points or the points where the forces act.*
The collision analysis is not a major proof in itself; but the perspective on collisions Would be improved with the correct definition of energy. The same is true of most points being made here. The mathematical proof is in applying rockets to the falling object issue, which is shown in the rocket section.
--------------------------------------------------------------------------------
*Notice in the rocket section that the analysis of power by erroneous concepts is made relative to the point where the forces act by using the separation velocity of the exhaust as the reference. This strange twist allows the rocket equations to be balanced with the erroneous definitions. It also contradicts the premise that energy addition must occur relative to an external reference frame by the erroneous concepts.
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ArchAngle
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Posted - 2004.07.08 08:19:00 -
[5]
Edited by: ArchAngle on 08/07/2004 08:27:34
Edited by: ArchAngle on 08/07/2004 08:22:51
Edited by: ArchAngle on 08/07/2004 08:20:09
pLEASE NOTE THAT THE TABLES DID NOT COME OUT CORECTLY SO IF YOU WISH TO USE THEN THERE IS 2 COLUMS PER VALUE FROM LEFT TO RIGHT AND 4 ROES
iF ANY ONE WILL HOST A COPY OF A HTML DOC ILL UP LOAD THE WHOLE THING INCLUDIGN THE TABLE
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ArchAngle
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Posted - 2004.07.08 08:19:00 -
[6]
Edited by: ArchAngle on 08/07/2004 08:27:34
Edited by: ArchAngle on 08/07/2004 08:22:51
Edited by: ArchAngle on 08/07/2004 08:20:09
pLEASE NOTE THAT THE TABLES DID NOT COME OUT CORECTLY SO IF YOU WISH TO USE THEN THERE IS 2 COLUMS PER VALUE FROM LEFT TO RIGHT AND 4 ROES
iF ANY ONE WILL HOST A COPY OF A HTML DOC ILL UP LOAD THE WHOLE THING INCLUDIGN THE TABLE
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ArchAngle
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Posted - 2004.07.08 09:00:00 -
[7]
Edited by: ArchAngle on 08/07/2004 09:03:01
Yes it would multipuly the damage 2x2=4 not 2x2=1 to put it simply Let me ask you this if you are driving a 2 tone truck and hit a GEO metro head on at 100mph and the other car is going at 100mph right at you what happens . the car is totald and you are probaly scaterd over 100 yards. not you hit the other car and it does less damage because you hit it head on. 
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ArchAngle
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Posted - 2004.07.08 09:00:00 -
[8]
Edited by: ArchAngle on 08/07/2004 09:03:01
Yes it would multipuly the damage 2x2=4 not 2x2=1 to put it simply Let me ask you this if you are driving a 2 tone truck and hit a GEO metro head on at 100mph and the other car is going at 100mph right at you what happens . the car is totald and you are probaly scaterd over 100 yards. not you hit the other car and it does less damage because you hit it head on.
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ArchAngle
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Posted - 2004.07.08 09:08:00 -
[9]
Edited by: ArchAngle on 08/07/2004 09:10:21
 Originally by: TomB
 Originally by: Imhotep Khem
Edited by: Imhotep Khem on 07/07/2004 15:36:51
Going to read some more of this thread before i make a final comment. I'm at a loss as to why a battleship should not take out a frigate with "2-3 cruise missiles" or 1-2 torpedoes."
Because an explosion of 400m3 shouldn't damage a frigate for 400m3
UM how is this true large the explosive payload of a missle wider area of effect kinda like 10 pounds of C4 will make a bigger boom than 1 ounce of C4 its all abpought the payload of the missle and how close the target is to the blast. bigger missle= bigger boom = more damage simple as that and if the frig is any were neer the blast it should take maxum damage not less because it s a small ship actualy it should take more damage because its a more fragile ship and has less armor to protect it
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ArchAngle
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Posted - 2004.07.08 09:08:00 -
[10]
Edited by: ArchAngle on 08/07/2004 09:10:21
Originally by: TomB
Originally by: Imhotep Khem
Edited by: Imhotep Khem on 07/07/2004 15:36:51
Going to read some more of this thread before i make a final comment. I'm at a loss as to why a battleship should not take out a frigate with "2-3 cruise missiles" or 1-2 torpedoes."
Because an explosion of 400m3 shouldn't damage a frigate for 400m3
UM how is this true large the explosive payload of a missle wider area of effect kinda like 10 pounds of C4 will make a bigger boom than 1 ounce of C4 its all abpought the payload of the missle and how close the target is to the blast. bigger missle= bigger boom = more damage simple as that and if the frig is any were neer the blast it should take maxum damage not less because it s a small ship actualy it should take more damage because its a more fragile ship and has less armor to protect it
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ArchAngle
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Posted - 2004.07.08 11:39:00 -
[11]
one quick question will there be a minum damage and a max damage so you still hit every time or is it a shoot all day long never hit crap nerf
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ArchAngle
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Posted - 2004.07.08 11:39:00 -
[12]
one quick question will there be a minum damage and a max damage so you still hit every time or is it a shoot all day long never hit crap nerf
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ArchAngle
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Posted - 2004.07.08 11:42:00 -
[13]
also its sooo anoying to cary 20 differnt amo types ina ship you ever try to change your load out based on ship type in a rapid senero or do it for several hours hunting rats its not fun maby make a multi purpous launcher again but make it with seige launcher requirments and base the fire time off the amo used
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ArchAngle
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Posted - 2004.07.08 11:42:00 -
[14]
also its sooo anoying to cary 20 differnt amo types ina ship you ever try to change your load out based on ship type in a rapid senero or do it for several hours hunting rats its not fun maby make a multi purpous launcher again but make it with seige launcher requirments and base the fire time off the amo used
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ArchAngle
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Posted - 2004.07.08 12:10:00 -
[15]
Edited by: ArchAngle on 08/07/2004 12:13:09
One of my primary concern is the way the game is being balanced its almost like there is no reward for palying long term its like everthing is being geared at noobs all experanced players are being penilised fro playing long time im proud of my skill gap i worked hard and invested many months in train my skills up higher. with all the changes its gettign harder and harder to be independant unless you want to be a carebear and hang in empire all day and mine. First it was guns now missles shure we get more powergride for guns but why do i even ned more power i got to use frig guns and launchers ona bs it wouldnt be so bad if there was a large launcher or gun thats built for anti frig. but as time goes on this is my perdiction of the future. im gona use a mega for my exaple a mega fitted with 4 425' 2 light nutrons one 150mm s rail and 1 cruse / seige launcher im not gona go in to the details of the mids and lows as they are not revelent here. now in your cargo hold you will have 250 rounds of antimater S for close range frig 200 rounds of iridum S for medum range frig duties 500 rounds of large antimater for closre range bs 300 rounds of iridum or uranium for mid range and 100-200 light em missles 100-200 thermal light missles or what ever type is aproit then on to large missles get my drift next thing you know you will have 500-600m3 of amo in your hold just to kill rats and have to switch all the types of amo on the fly every time you start a new spawn and evey time they change range and even if you by the grace of god get the rats killed where is the loot gona go as you have no room because of the amo get my drift and even if you only change amo when absulty nessary you will still have to change every gun 3-4 times a fight jut to hit desent damage even if you even hit.90% of you amo gets blown in to space.lets see here in a hour you kill 3-4 spawns of varign types you spend 15 min per spawn so that means you have reloaded your guns 12 times at least x7 guns thast abought 2 minuts from jut changing amo types not to mention reloadign your guns and launchers i think if made my point its great that your giving ships a purpous but making us haul around a lot of differnt types of amo and missles and fit a bunch of differnt guns is crazy why not just lock the gun type to the ship and make new wepons for each task or adapt the low end guns for anti frig and cruser so we can cary one sise of amo anf a multi purpious launcher would be nice depending on ship class its just my thoughts cause it funny to be fitted for anti frig kill the frigs and get raped by the bs's or kill the bs's and get raped by frigs. do fig
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ArchAngle
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Posted - 2004.07.08 12:10:00 -
[16]
Edited by: ArchAngle on 08/07/2004 12:13:09
One of my primary concern is the way the game is being balanced its almost like there is no reward for palying long term its like everthing is being geared at noobs all experanced players are being penilised fro playing long time im proud of my skill gap i worked hard and invested many months in train my skills up higher. with all the changes its gettign harder and harder to be independant unless you want to be a carebear and hang in empire all day and mine. First it was guns now missles shure we get more powergride for guns but why do i even ned more power i got to use frig guns and launchers ona bs it wouldnt be so bad if there was a large launcher or gun thats built for anti frig. but as time goes on this is my perdiction of the future. im gona use a mega for my exaple a mega fitted with 4 425' 2 light nutrons one 150mm s rail and 1 cruse / seige launcher im not gona go in to the details of the mids and lows as they are not revelent here. now in your cargo hold you will have 250 rounds of antimater S for close range frig 200 rounds of iridum S for medum range frig duties 500 rounds of large antimater for closre range bs 300 rounds of iridum or uranium for mid range and 100-200 light em missles 100-200 thermal light missles or what ever type is aproit then on to large missles get my drift next thing you know you will have 500-600m3 of amo in your hold just to kill rats and have to switch all the types of amo on the fly every time you start a new spawn and evey time they change range and even if you by the grace of god get the rats killed where is the loot gona go as you have no room because of the amo get my drift and even if you only change amo when absulty nessary you will still have to change every gun 3-4 times a fight jut to hit desent damage even if you even hit.90% of you amo gets blown in to space.lets see here in a hour you kill 3-4 spawns of varign types you spend 15 min per spawn so that means you have reloaded your guns 12 times at least x7 guns thast abought 2 minuts from jut changing amo types not to mention reloadign your guns and launchers i think if made my point its great that your giving ships a purpous but making us haul around a lot of differnt types of amo and missles and fit a bunch of differnt guns is crazy why not just lock the gun type to the ship and make new wepons for each task or adapt the low end guns for anti frig and cruser so we can cary one sise of amo anf a multi purpious launcher would be nice depending on ship class its just my thoughts cause it funny to be fitted for anti frig kill the frigs and get raped by the bs's or kill the bs's and get raped by frigs. do fig
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ArchAngle
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Posted - 2004.07.08 13:19:00 -
[17]
i dont so much mine the changes but the fact that you have to caryt 10 differnt kinds of amo around to kill soem npc is crazy im looking at it in a hunting perpective i dont mine it for anti frig or somthign in pvp but hunting npc its just anoying and hunting in a cruser is imposable solo all i ask for is large guns for antii frig and proper guns for the ship class so i dont need small or medum guns on y bs i save them for my cruer and frigs
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ArchAngle
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Posted - 2004.07.08 13:19:00 -
[18]
i dont so much mine the changes but the fact that you have to caryt 10 differnt kinds of amo around to kill soem npc is crazy im looking at it in a hunting perpective i dont mine it for anti frig or somthign in pvp but hunting npc its just anoying and hunting in a cruser is imposable solo all i ask for is large guns for antii frig and proper guns for the ship class so i dont need small or medum guns on y bs i save them for my cruer and frigs
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ArchAngle
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Posted - 2004.07.09 20:38:00 -
[19]
Originally by: Grut
I dont like the idea of frigs & cruisers taking dmg from missiles at high speed, large guns give a miss so why shouldnt missiles?
Maybe some kind of trigger time would work, based on signal radius and combined with what you said about explosion radius earlier would work.
Something like;
UM heck no First of all a frig is a 100kisk peice of hardwear and as far as im concerned if they move fast enough they can get away from missles but for a frig to be neerly invincable is stupid why fly a high end ship at all lets all just fly frigs and go back to beta make things simple or beter yet do away with missles guns and npc's and well just all mine for ever. basicly atm missle physics is preety acrut (if you can out manver the missle you dotn get hurt ) maby just add soem missle counter measures to frigs maby a bit more agilty and maby a missle evasion skill or somthing or electronic missle jamming abiltys so the modual messes with t the missle logic and the missle misses or flyes off in to space. why change the whole physics system for missles as they are preety acrut atm.
torp trigger time 10000ms
torp explosion radius 400
frig signal radius 100
trigger time 10000/100 = 1 second, the frig has 1 second to move 400m to avoid being hit, if not it takes a fraction of the blast as previously described.
Light missiles would have alot smaller trigger time.
If your using signal radius for the above you could knock it out of later calcs, of course this would mean stationary frigs get hit by the full whack of dmg  but imo thats a nice counter to missiles not being able to hit frigs at range like guns.
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ArchAngle
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Posted - 2004.07.09 20:38:00 -
[20]
Originally by: Grut
I dont like the idea of frigs & cruisers taking dmg from missiles at high speed, large guns give a miss so why shouldnt missiles?
Maybe some kind of trigger time would work, based on signal radius and combined with what you said about explosion radius earlier would work.
Something like;
UM heck no First of all a frig is a 100kisk peice of hardwear and as far as im concerned if they move fast enough they can get away from missles but for a frig to be neerly invincable is stupid why fly a high end ship at all lets all just fly frigs and go back to beta make things simple or beter yet do away with missles guns and npc's and well just all mine for ever. basicly atm missle physics is preety acrut (if you can out manver the missle you dotn get hurt ) maby just add soem missle counter measures to frigs maby a bit more agilty and maby a missle evasion skill or somthing or electronic missle jamming abiltys so the modual messes with t the missle logic and the missle misses or flyes off in to space. why change the whole physics system for missles as they are preety acrut atm.
torp trigger time 10000ms
torp explosion radius 400
frig signal radius 100
trigger time 10000/100 = 1 second, the frig has 1 second to move 400m to avoid being hit, if not it takes a fraction of the blast as previously described.
Light missiles would have alot smaller trigger time.
If your using signal radius for the above you could knock it out of later calcs, of course this would mean stationary frigs get hit by the full whack of dmg but imo thats a nice counter to missiles not being able to hit frigs at range like guns.
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ArchAngle
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Posted - 2004.07.09 20:42:00 -
[21]
Also a simple soultion for the raven is to remove the gun points from the raven and add 2 more launcher slots
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ArchAngle
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Posted - 2004.07.09 20:42:00 -
[22]
Also a simple soultion for the raven is to remove the gun points from the raven and add 2 more launcher slots
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ArchAngle
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Posted - 2004.07.14 05:20:00 -
[23]
Originally by: Aequitas Veritas
Originally by: ArchAngle
Also a simple soultion for the raven is to remove the gun points from the raven and add 2 more launcher slots
Woudl probably make it too strong dont you think? :) at least if it can fit 8 siege launchers
Well my thought is leave the power grid and cpu thesame forcing you to use standard or hevy launchers
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ArchAngle
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Posted - 2004.07.14 05:20:00 -
[24]
Originally by: Aequitas Veritas
Originally by: ArchAngle
Also a simple soultion for the raven is to remove the gun points from the raven and add 2 more launcher slots
Woudl probably make it too strong dont you think? :) at least if it can fit 8 siege launchers
Well my thought is leave the power grid and cpu thesame forcing you to use standard or hevy launchers
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