Difference between revisions of "Generic EM Drive Information"

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(Velocity: Added in a section about Kinetic Energy.)
m (More information on Kinetic energy.)
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*deltaKineticEnergy = (1/2)*Mass*((delta velocity)^2)
 
*deltaKineticEnergy = (1/2)*Mass*((delta velocity)^2)
 
*deltaKineticEnergy = (1/2)*Mass*([(ThrustForce/Mass)*(delta time) ]^2)
 
*deltaKineticEnergy = (1/2)*Mass*([(ThrustForce/Mass)*(delta time) ]^2)
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For the EM Drive to make any sense as a '''closed system''' it implies having memory of its time-history in order to never exceed the critical velocity at which (for constant InputPower throughout the time period (delta time)):
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*InputPower*(delta time) = (1/2)*Mass*[(ThrustForce/Mass)*(delta time) ]^2 <ref>Consistent with equation 19-1: Rocket Propulsion Elements 7th edition- Sutton</ref>
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If the EM drive conserves momentum by interacting with the QuantumVacuum for example, we have an '''open system''' where momentum will be flowing in or out of the cavity, and therefore the open system should be taken into account in the conservation of kinetic energy (which we have not done) in order to arrive at a satisfactory answer.
  
 
== References ==
 
== References ==
 
<references />
 
<references />

Revision as of 12:53, 14 May 2015

This is a location for assorted information concerning EM Drives. Likely applies to all current experimental efforts.

Velocity

At constant input power, the thrust, and therefore the acceleration, must decrease with time, to ensure that the spacecraft's (change in) velocity never exceeds 2*Power/Thrust.[1] Effectively, White proposes that the maximum velocity possibly achievable for these EM Drives is 2*PowerInput/ThrustForce. This energy constraint is known even from Sutton's textbook on Elements of Rocket Propulsion (equation 19-1 Rocket propulsion elements 7th edition- Sutton).

The higher the ThrustForce/InputPower, the lower the maximum velocity of an EM Drive: MAXIMUM VELOCITY OF EM DRIVE = 2 /(thrustForce/PowerInput)

  1. (Brady TE mode) ThrustForce/power: 0.00002131 Newton/Watt --> MaximumVelocity = 2/0.00002131 m/s = 93853 m/s = 93.85 km/s
  2. (Prof. Juan Yang et.al. China) ThrustForce/power: 0.000290 Newton/Watt --> MaximumVelocity = 2/0.000290 m/s = 6897 m/s = 6.90 km/s
  3. (Cannae Superconducting) ThrustForce/power: 0.0009524 Newton/Watt --> MaximumVelocity = 2/0.0009524 m/s = 2100 m/s = 2.10 km/s

The maximum velocity of the EM Drive is inversely proportional to Q: [2] Therefor the higher the Q, the lower the maximum velocity.

  1. MAXIMUM VELOCITY OF EM DRIVE = 2 /(thrustForce/PowerInput)
  2. MAXIMUM VELOCITY OF EM DRIVE ~ (2 /(Q*otherParameters))

Kinetic Energy

The frame-of-reference kinetic energy issues (for non-relativistic spacecraft speeds) are eliminated by the definition of constant acceleration as acceleration=(delta velocity)/(delta time) so that (delta velocity)=acceleration*(delta time).[3] Since from Newton's 2nd law, acceleration=ThrustForce/Mass, the change in velocity is uniquely defined in terms of: ThrustForce, total Mass of the spacecraft and DeltaTime (regardless of velocity-frame-of-reference) directly as (delta velocity)=(ThrustForce/Mass)*(delta time). The intrinsic definition of the change in KineticEnergy of the spacecraft under constant acceleration:

  • deltaKineticEnergy = (1/2)*Mass*((delta velocity)^2)
  • deltaKineticEnergy = (1/2)*Mass*([(ThrustForce/Mass)*(delta time) ]^2)

For the EM Drive to make any sense as a closed system it implies having memory of its time-history in order to never exceed the critical velocity at which (for constant InputPower throughout the time period (delta time)):

  • InputPower*(delta time) = (1/2)*Mass*[(ThrustForce/Mass)*(delta time) ]^2 [4]

If the EM drive conserves momentum by interacting with the QuantumVacuum for example, we have an open system where momentum will be flowing in or out of the cavity, and therefore the open system should be taken into account in the conservation of kinetic energy (which we have not done) in order to arrive at a satisfactory answer.

References