In reply to David Roberson's message of Fri, 28 Nov 2014 16:23:23 -0500: Hi Dave, [snip] >Kinetic energy is calculated directly by the magnitude of relative velocity of >the ship to the observer and therefore the second guy sees essentially no net >change in kinetic energy once the drive cycle is completed.
>Also notice that the energy is nonlinear with velocity. The fact that it is >proportional to the second power of the velocity allows the direction to >become unimportant. > >I am beginning to suspect that you are playing games at this point. All you >need to do is to look up the definition of kinetic energy to see that what I >am stating is correct. I have no idea why you think kinetic energy has >anything to do with the change in velocity instead of the net relative >velocity. Velocity is a vector quantity, which means that it has both direction and magnitude. Before you square the quantity, you need to do vector addition (or subtraction in this case). The fact that the direction has changed, makes a difference in the change in kinetic energy. If you prefer, consider the fact that in order to change direction, the craft has to first decelerate until its speed is zero, then accelerate again in the opposite direction. Both observers see the same acceleration of the craft, but for one of them this results in a velocity of 2 m/s in one direction, while for other it appears as a change in velocity from +1 m/s to -1 m/s (or if you prefer from -1 m/s to +1 m/s). > Perhaps you can quote a source. ;) Any physics book. > >Dave > > > > > > > >-----Original Message----- >From: mixent <mix...@bigpond.com> >To: vortex-l <vortex-l@eskimo.com> >Sent: Fri, Nov 28, 2014 4:02 pm >Subject: Re: [Vo]:They call me a moron. A reply. > > >In reply to David Roberson's message of Thu, 27 Nov 2014 23:47:17 -0500: >Hi Dave, >[snip] > >The "v" in the formula Ek = 1/2 mv^2 actually applies to the change in >velocity, >not velocity in any absolute sense. For the sake of convenience, we normally >choose a frame of reference in which the initial velocity is zero which makes >the calculation simpler. > >For observer 1, the change is 1/2 * m * (2-0)^2 = 1/2 * m * (2)^2. >For observer 2, the change is 1/2 * m * (1 - -1)^2 = 1/2 * m * (2)^2. > >I.e. they both see the same change in kinetic energy. > >Note 1: I have not included the dimensions here to keep the formula as simple >as >possible in ASCII text. >Note 2: Depending on the initial direction of motion, you may choose to write >the equation for observer 2 as 1/2 * m * (-1 - 1)^2 = 1/2 * m * (-2)^2, however >this still gives the same result for the kinetic energy change. > >> >> Robin, I just came up with a thought experiment that lends support to the >> idea >that a reactionless drive is not likely to exist. Take 2 different observers, >one that is moving beside the ship at the same velocity as it has prior to >activating the drive. The second one is moving at a velocity that allows him >to >observe the ship decelerate first until it reaches a velocity of zero relative >to him and then to accelerate in the reverse direction until it reaches the >exact same original velocity in the opposite direction. >> >>The first observer sees the velocity of the ship go from for this example 0 >meters per second to 2 meters per second. He determines that the ship now has >2*2*Mass/2 units of kinetic energy. The amount of internal mass that the ship >burns up to achieve this acceleration is extremely small and can almost be >neglected. >> >>The second observer sees the ship moving at the same speed before and then >after the application of the drive. The only difference he measures is that >the direction of the motion of the ship is reversed by the drive. So he sees >the ship begin the motion moving 1 meter per second relative to him initially >and then after the drive shuts down the ship is moving 1 meter per second in >the opposite direction. This observer determines that the kinetic energy of >the ship has not changed measurably due to the application of the drive. >> >>At this low velocity the second observer determines that the mass converted >into drive power is essentially the same as that determined by the first >observer. Both guys have a very hard time figuring out exactly how much mass >is >converted, and they agree that any difference is hidden in the noise. >> >>In this experiment we have two independent observers seeing the same ship >>being >subject to the same drive. The amount of kinetic energy being deposited to >the >ship by essentially the same loss of internal mass varies remarkably according >to each. This does not add up. >> >>As I have mentioned before, with a normal drive this case can be handled >without a problem. The exhaust material supplies the kinetic energy and >momentum needed to balance the equation. >> >>Apparently every observer moving at a different initial velocity relative to >the driven ship arrives at a significantly different calculation regarding the >ships energy balance after the application of the drive. How could something >this radical be possible? Let me say it again, there is no problem of this >sort >to deal with when a standard drive is applied. >> >>Dave >> >Regards, > >Robin van Spaandonk > >http://rvanspaa.freehostia.com/project.html > > > Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
