vuquta Posted January 26, 2010 Author Posted January 26, 2010 Originally Posted by vuquta Given any time interval, A and B are at some positions in M' and M' calculates that interval as ∆t and views that interval as ∆t/λ elapsed in M. And vice versa. You can't just set up a one to one correspondence between reference frames. Each frame views the other as under time compression. Sure I can set up a correspondence. Let clocks of M' be synchronized. Each clock of M becomes a coordinate system. Set x = vt for the moving system, and t' = ( t - vx/c² )λ Then you get, t' = t/λ for any interval. M' is not concluding M has different times at different positions of M unless light receiving events are taking place. This is GPS. The earth frame does not conclude two satellites in the same orbit have different times based on positions. I have no idea what this section means. Yea, I see crappy writing. I meant that because of GPS, the earth frame cannot conclude GPS satellites are not in sync based on position. This is supported by the GPS system. Thus, if two satellites emitted light at exactly 12:00 UTC, the earth frame would not translate that to two different times. Thus, the earth frame would conclude the emissions were simultaneous. So, when the interval is at ∆t1/λ in M, both A and B emit. In M? As viewed by who? This is from the view of M. That corresponds to one interval ∆t1 in M'. No, it doesn't. There isn't a direct correspondence like that. Yes, there is and this is supported by GPS otherwise, the system would fail. That is because an earth clock would have to conclude a GPS1 and GPS2 did not have the same time and any time t in its frame if they are at different positions according to this branch of SR logic. This is not supported by the scientific evidence.
swansont Posted January 26, 2010 Posted January 26, 2010 M' is co-located with M when both light beams meet at M. There is no choice that M' is also met by both light beams simultaneously also. No, you are merely insisting on that —*you have to show it's true without assuming anything about the simultaneity. Relativity says it's wrong, so you won't be able to do this with the Lorentz transforms. The only way to prove this would be with a physical experiment.
vuquta Posted January 26, 2010 Author Posted January 26, 2010 No, you are merely insisting on that —*you have to show it's true without assuming anything about the simultaneity. Relativity says it's wrong, so you won't be able to do this with the Lorentz transforms. The only way to prove this would be with a physical experiment. Good argument. Actually, SR says that B will emit before A from the view of M' and M concludes they emit simultaneously. From this, SR claims the clock of B in M' is ahead of the clock of A. Thus, from the view of M', the "emission time t0" in M is reached at B first so emission occurs and then eventually that t0 is reached at A and emission occurs. Now, let M and M' agree on some start time. Let M' be an array of synched clocks. A corrsponding clock that is located with B and any time t, would claim the clock at B is beating faster than the M' clock. Also, handling the A clock the same way, the M' clock would record that the clock of A is beating slower than the M' time. Anyway, GPS does not beat at different rates based on satellite positions. So, it is my view that GPS is the experiment. What is your opinion?
Sisyphus Posted January 26, 2010 Posted January 26, 2010 (edited) Anyway, GPS does not beat at different rates based on satellite positions. So, it is my view that GPS is the experiment. I still don't understand what you mean by this. GPS satellites do compensate for relativistic effects: http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html And that's hardly the only "experiment:" http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html If you think that special relativity is incorrect, what is your view on the fact that it makes accurate predictions of observed reality, and you do not? Edited January 26, 2010 by Sisyphus
swansont Posted January 26, 2010 Posted January 26, 2010 Good argument. Actually, SR says that B will emit before A from the view of M' and M concludes they emit simultaneously. From this, SR claims the clock of B in M' is ahead of the clock of A. Thus, from the view of M', the "emission time t0" in M is reached at B first so emission occurs and then eventually that t0 is reached at A and emission occurs. Now, let M and M' agree on some start time. Let M' be an array of synched clocks. A corrsponding clock that is located with B and any time t, would claim the clock at B is beating faster than the M' clock. Also, handling the A clock the same way, the M' clock would record that the clock of A is beating slower than the M' time. And if M' concludes that the emission is not simultaneous, then the light will not reach M' simultaneously. And if you bring clocks into the problem, you will conclude that the timing doesn't agree, consistent with the disagreement in simultaneity. Anyway, GPS does not beat at different rates based on satellite positions. So, it is my view that GPS is the experiment. What is your opinion? You are ignoring that satellite clocks are not identical to the ground clocks, so you can't make a direct comparison to the train problem, and your analysis of the train problem is flawed, and I don't see the utility of applying a flawed analysis to GPS. Merged post follows: Consecutive posts mergedSure I can set up a correspondence. Let clocks of M' be synchronized. Each clock of M becomes a coordinate system. Set x = vt for the moving system, and t' = ( t - vx/c² )λ Then you get, t' = t/λ for any interval. Simple math problem: If x = vt and t' = ( t - vx/c² )λ The only way to get t' = t/λ is for x=0 Since v ≠ 0, I don't see how this is possible "for any interval." It can only be true at one particular value of t (and t'), i.e. the time when the two points are co-located.
vuquta Posted January 26, 2010 Author Posted January 26, 2010 (edited) I still don't understand what you mean by this. GPS satellites do compensate for relativistic effects: http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html I mean, ignoring GR and Sagnac since they are not part of this problem, the GPS system preprograms the satellite clock according to time dilation prior to launch. All satellites in one orbit frame have one time. This is also true from the view of the ground clocks. So, if two GPS satellites emitted lasers at UTC 12:00, then the ground would conclude the emissions were simultaneous. R of S needs emissions not to be simultaneous to work. Now, I am not talking reception of the light. That is very clear. If a stationary frame sees two equal light path lengths, then a moving frame will record two different light path lengths. We are not talking light paths though. We are talking emission times. And that's hardly the only "experiment:" http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html There is nothing in here to address this problem. One would need to perform one way speed of light calculations using clocks. None have been done. If you think that special relativity is incorrect, what is your view on the fact that it makes accurate predictions of observed reality, and you do not? What predictions have I made that are not accurate? Merged post follows: Consecutive posts mergedAnd if M' concludes that the emission is not simultaneous, then the light will not reach M' simultaneously. And if you bring clocks into the problem, you will conclude that the timing doesn't agree, consistent with the disagreement in simultaneity. if M' concludes that the emission is not simultaneous, then the light will not reach M' simultaneously This may or may not be true. M contends the emission are simultaneous. If they are also simultaneous emissions in M', which is consistent with time dilation, then M' is struck simultaneously while co-located with M and therefore, M' must conclude two different light speeds. You must bring timing into this experiment also. That is the nature of R of S. My contentions is that clocks frame to frame only beat time dilated. R of S requires them to beat differently at different positions. You are ignoring that satellite clocks are not identical to the ground clocks, so you can't make a direct comparison to the train problem, and your analysis of the train problem is flawed, and I don't see the utility of applying a flawed analysis to GPS. You are ignoring that satellite clocks are not identical to the ground clocks, so you can't make a direct comparison to the train problem, They are identical except they are corrected for GR, SR and sagnac. Leave in GR and sagnac corrections because they are not part of this problem and therefore that removes their effects as if they did not exist, they all you have is the orbital frame beating time dilation and synchronous. This is consistent with my reasoing but not with R of S. Simple math problem: If x = vt and t' = ( t - vx/c² )λ The only way to get t' = t/λ is for x=0 Since v ≠ 0, I don't see how this is possible "for any interval." It can only be true at one particular value of t (and t'), i.e. the time when the two points are co-located. That is the incorrect way to derive time dilation. Here is the correct way. x = vt and t' = ( t - vx/c² )λ substitute x=vt because we are looking at a moving clock along the x-axis at any time t. t' = ( t - v(vt)/c² )λ t' = t( 1 - v²/c² )λ t' = t( (c² - v²) /c² )λ t' = t( (c² - v²) /c² )( √[c²/(c² - v²)] ) t' = t( √[(c² - v²) /c²] ) t' = t/λ. Thus, the moving t' clock beats slower than the stationary clock. You can find this analysis by Einstein in the middle of sections 4. http://www.fourmilab.ch/etexts/einstein/specrel/www/ ------------------------------------------------------------------------------------------------------------------- OK, here is the bottom line on what this thought experiment exposes. If you assume time dilation is true, then all clocks in a moving frame beat the same time dilated value compared to a corresponding clock in the stationary frame. So, if O is a line of synchronized clocks that is stationary, and O' is a line of synchronized clocks that is moving, any query of a stationary clock to a moving clock for an interval t, will find that moving clock to read t/λ. If all points of O query simultaneously, then all will conclude the same result. This is time dilation. Proof Given the description of O above, let cl be any clock in O' and place an O origin at that clock location. That clock will have a position x at after time t according to x = vt after the start time. Also, t' = ( t - vx/c² )λ. Substitute x = vt. t' = ( t - v(vt)/c² )λ. t' = t( 1 - v² /c² )λ t' = t( 1 - v² /c² )λ. t' = t( (c² - v²) /c² )λ. t' = t( (c² - v²) /c² )√[c²/(c² - v²)] t' = t √[(c² - v²)/c²] t' = t / λ. Since the point was chosen arbitrarily, this is true for all points in O'. Since t was used in the equation and this is the time of all synchronized clocks in the O frame, then the proof is complete. Now, if one assumes time dilation is true, then if a frame M emits light at 2 different points simultaneously, then a moving frame M' will conclude the emissions were simultaneous but at the same time dilated interval for both. Since the light paths are different lengths, M' is struck simultaneously for both light beams and both emitted simultaneously, then M' will conclude t=d/c for both paths is false. Other the other hand, if you assume t=d/c for both paths is true, then you must conclude point B emits before point A. But, this contradicts time dilation. Now, GPS and other experiments support time dilation. So that assumption has been validated. However, no one way time light timing experiments have been performed using clocks to support the other assumption that t=d/c is universal. Either way, both cannot be true. Edited January 26, 2010 by vuquta Consecutive posts merged.
swansont Posted January 27, 2010 Posted January 27, 2010 This may or may not be true. M contends the emission are simultaneous. If they are also simultaneous emissions in M', which is consistent with time dilation, then M' is struck simultaneously while co-located with M and therefore, M' must conclude two different light speeds. But they are not simultaneous in M'. The analysis shows that they are not. So the whole "if" is a circular argument. There is nothing else here. You must bring timing into this experiment also. That is the nature of R of S. My contentions is that clocks frame to frame only beat time dilated. R of S requires them to beat differently at different positions. No, relativity requires them to beat differently (frequency) at different speeds. It also requires them to display different times at different positions. Because phase accumulates as the position changes, because time has passed. They are identical except they are corrected for GR, SR and sagnac. Leave in GR and sagnac corrections because they are not part of this problem and therefore that removes their effects as if they did not exist, they all you have is the orbital frame beating time dilation and synchronous. This is consistent with my reasoing but not with R of S. "Correcting for SR" means that they are not identical clocks as with the train analysis. They run at a different frequency when in the same frame. You are comparing two different situations, and it's not valid. That is the incorrect way to derive time dilation. Here is the correct way. x = vt and t' = ( t - vx/c² )λ substitute x=vt because we are looking at a moving clock along the x-axis at any time t. t' = ( t - v(vt)/c² )λ t' = t( 1 - v²/c² )λ t' = t( (c² - v²) /c² )λ t' = t( (c² - v²) /c² )( √[c²/(c² - v²)] ) t' = t( √[(c² - v²) /c²] ) t' = t/λ. Thus, the moving t' clock beats slower than the stationary clock. But that's not true over an interval. That's true at a particular time (or position), and contradicts your previous contention that the second term somehow isn't included in a time dilation calculation. (i.e. the nonsense about separate dilation and simultaneity terms) You just showed that the whole equation reduces to t' = t/λ when you pick a final position (or time).
vuquta Posted January 27, 2010 Author Posted January 27, 2010 (edited) But they are not simultaneous in M'. The analysis shows that they are not. So the whole "if" is a circular argument. There is nothing else here. Well, I gave both arguments and you are asserting the one that refutes time dilation and has been experimentally verified. No, relativity requires them to beat differently (frequency) at different speeds. It also requires them to display different times at different positions. Because phase accumulates as the position changes, because time has passed. No, time dilation does not require this and this contradicts time dilation. Further this does not show up in GPS. In addition I am backed up by the scientific evidence of time dilation that does not depend on position and is simply consistent within the frame. Tests of Time Dilation and Transverse Doppler Effect http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Tests_of_time_dilation "Correcting for SR" means that they are not identical clocks as with the train analysis. They run at a different frequency when in the same frame. You are comparing two different situations, and it's not valid. False, they are corrected for time dilation. But that's not true over an interval. That's true at a particular time (or position), and contradicts your previous contention that the second term somehow isn't included in a time dilation calculation. (i.e. the nonsense about separate dilation and simultaneity terms) You just showed that the whole equation reduces to t' = t/λ when you pick a final position (or time). That whole "nonsense" was to try to help you understand that the shift term does not apply for clocks. You disagreed with it then but now agree with it. You must refute time dilation to maintain your position or redefine it. Let's see your proof against mine that time dilation is different for each clock in a moving frame. I am in the mainstream with experiments to back up my logic. Edited January 27, 2010 by swansont fix quote tag
swansont Posted January 27, 2010 Posted January 27, 2010 No, time dilation does not require this and this contradicts time dilation. Further this does not show up in GPS. In addition I am backed up by the scientific evidence of time dilation that does not depend on position and is simply consistent within the frame. It's quite apparent that you are not picking up on some of the the subtleties here. I did not say that time dilation depends on position, I said it depends on speed. (for you to baldly assert that it does not is absurd. The Gamov factor is a function of v) What I said was that for a moving clock, the phase depends on position. This is because position depends on time, and as time passes, the moving clock will accumulate more phase, but at a reduced rate because of dilation. If λ=2, after 1 second of time in the stationary frame, the moving clock will gave traveled some distance d, and the clock will read 1/2 second. After 2 seconds, the moving clock will have moved 2d, and the clock will read 1 second. So the displayed time on the clock (which is the phase) depends on x and t. The rate (frequency) depends on v. And yes, GPS accounts for this. Again, your bald assertion to the contrary is absurd. http://relativity.livingreviews.org/Articles/lrr-2003-1/ Look at equation 28. it gets absorbed into another term in equation 33 v^2/2 is the kinematic time dilation if v<<c . Do a binary expansion on the Gamov factor and see for yourself False, they are corrected for time dilation. In GPS they are. In the train experiment they are not. Which is why you cannot blindly use the analysis of one and apply it to the other — you need to account for the fact that the clocks are not identical in any GPS analysis. That whole "nonsense" was to try to help you understand that the shift term does not apply for clocks. You disagreed with it then but now agree with it. You must refute time dilation to maintain your position or redefine it. Let's see your proof against mine that time dilation is different for each clock in a moving frame. When you did your analysis, you used both terms in the time dilation formula to get t' = t/λ. Which confirms that the entire equation is the formula for time dilation. I am in the mainstream with experiments to back up my logic. Since you insist that GPS analysis is wrong, I hardly think this is the case.
vuquta Posted January 27, 2010 Author Posted January 27, 2010 It's quite apparent that you are not picking up on some of the the subtleties here. I did not say that time dilation depends on position, I said it depends on speed. (for you to baldly assert that it does not is absurd. The Gamov factor is a function of v) What I said was that for a moving clock, the phase depends on position. This is because position depends on time, and as time passes, the moving clock will accumulate more phase, but at a reduced rate because of dilation. If λ=2, after 1 second of time in the stationary frame, the moving clock will gave traveled some distance d, and the clock will read 1/2 second. After 2 seconds, the moving clock will have moved 2d, and the clock will read 1 second. So the displayed time on the clock (which is the phase) depends on x and t. The rate (frequency) depends on v. And yes, GPS accounts for this. Again, your bald assertion to the contrary is absurd. http://relativity.livingreviews.org/Articles/lrr-2003-1/ Look at equation 28. it gets absorbed into another term in equation 33 v^2/2 is the kinematic time dilation if v<<c . Do a binary expansion on the Gamov factor and see for yourself Well, you are intelligent. We are not communicating. I never said time dilation did not depend on v. That indeed would be absurd. I said it did not depend on position. A moving clock at position A1 moving to B1 will experience the same time interval as a clock at a different position A2 moving to B2 givemn the same moving frame. Einstein: It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be second slow. Thence we conclude that a balance-clock7 at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions. Now, he said time dilation applies to the time interval for a clock to move from point A to to B. Since the position A is arbitrary, then all clocks in a moving frame beat in sync at a time dilated interval from the view of the stationary frame since A can be anwhere in the frame. Let's be specific. From the earth frame, are the satellite clocks in sync or not. In GPS they are. In the train experiment they are not. Which is why you cannot blindly use the analysis of one and apply it to the other — you need to account for the fact that the clocks are not identical in any GPS analysis. Uh, they are the same clocks. The same clocks only means they are able to maintain the same frequency and about the same accuracy. When you did your analysis, you used both terms in the time dilation formula to get t' = t/λ. Which confirms that the entire equation is the formula for time dilation. That is correct and I think you are getting it. When using clocks as moving objects, you use time dilation. When looking at light moving, you use full LT. Thus, if clocks in a moving frame are synched at some agreed upon time, then they remain in sync except they will all be time dilated for the interval of time of the stationary frame. This contradicts t=d/c as universal. Since you insist that GPS analysis is wrong, I hardly think this is the case. No, I am OK with the GPS.
swansont Posted January 28, 2010 Posted January 28, 2010 Uh, they are the same clocks. The same clocks only means they are able to maintain the same frequency and about the same accuracy. No, the frequency if the GPS clocks are adjusted to be different on the ground, so that they will compensate for the gravitational and kinematic effects when in orbit. Identical clocks run at the same frequency when in the same reference frame. Since the clocks are not identical, you cannot apply an analysis that depends on them being identical.
vuquta Posted January 28, 2010 Author Posted January 28, 2010 No, the frequency if the GPS clocks are adjusted to be different on the ground, so that they will compensate for the gravitational and kinematic effects when in orbit. Identical clocks run at the same frequency when in the same reference frame. Since the clocks are not identical, you cannot apply an analysis that depends on them being identical. Does not work. These adjustments remove the effects. So, your view is simply not working. When you do an experiment, you want to exclude other factors, no? Try again. Time dilation refutes the universal t=d/c.
swansont Posted January 28, 2010 Posted January 28, 2010 Does not work. These adjustments remove the effects. So, your view is simply not working. When you do an experiment, you want to exclude other factors, no? Try again. Time dilation refutes the universal t=d/c. No, the adjustments compensate for the effects. From a physics standpoint, that's a huge difference. You must still analyze the situation properly, and you aren't doing that. BTW, GPS works. Which kinda means my view works.
vuquta Posted January 28, 2010 Author Posted January 28, 2010 No, the adjustments compensate for the effects. From a physics standpoint, that's a huge difference. You must still analyze the situation properly, and you aren't doing that. BTW, GPS works. Which kinda means my view works. Well, my argument agrees that GPS works. Each satellite remains synched within 100 nano-seconds of the ground clocks. GPS Specification States +/- 100 Nanoseconds to UTC http://www.endruntechnologies.com/timingspec.htm So, if GPS1 emits a laser sat 12:00 UTC and GPS2 emits at 12:00 UTC, then the ground clocks all agree the laser emits ~ at the same time. They are not out of sync up to the limits of the technology. Now, that means in my thought experiment, the M frame A and B emits lasers simultaneously, thus the M' frame will conclude they are simultaneous and thus will measure a different value c for each direction. This contradicts SR.
swansont Posted January 28, 2010 Posted January 28, 2010 100 nanoseconds is an operational requirement, and not a ramification of any physics. The ground clocks are synched up to GPS with a much tighter tolerance, it's just that the ground clocks may not reflect UTC that precisely. This has ramifications for people who want to get UTC from GPS clocks. IOW, UTC(GPS) - UTC(USNO) is smaller than 100 nanoseconds. This says nothing about the difference between the ground clocks for GPS and the satellite clocks. As far as the rest: Your 1-D analysis was shown to be flawed; you assumed a condition of simultaneity that is not part of (and is contradicted by) relativity. If you apply this to GPS, the analysis of that will be similarly flawed. I'm done here. There is no new ground to go over. You are merely repeating the same thing over and over, and repetition does not make things right.
vuquta Posted January 28, 2010 Author Posted January 28, 2010 100 nanoseconds is an operational requirement, and not a ramification of any physics. The ground clocks are synched up to GPS with a much tighter tolerance, it's just that the ground clocks may not reflect UTC that precisely. This has ramifications for people who want to get UTC from GPS clocks. IOW, UTC(GPS) - UTC(USNO) is smaller than 100 nanoseconds. This says nothing about the difference between the ground clocks for GPS and the satellite clocks. As far as the rest: Your 1-D analysis was shown to be flawed; you assumed a condition of simultaneity that is not part of (and is contradicted by) relativity. If you apply this to GPS, the analysis of that will be similarly flawed. I'm done here. There is no new ground to go over. You are merely repeating the same thing over and over, and repetition does not make things right. OK, you believe the satellites are not in sync from the view of the ground clocks. You believe also the emission are not in sync from M' and thus M' will believe B was emitted before A in order to preserve SR. This differential is caused by the distance to M' upon light emission. You believe they are not in sync based on distance to the clock. This is not supported by GPS but that is OK. If you read here, you will find the triangulation depends on the satellites have the same common time and earth time and then use distance to determine position. If the satellites were not in sync, this would not work. So, you are wrong in your thesis. http://www.beaglesoft.com/gpstechnology.htm Now, going back to the original R of S thought experiment from SR, both clocks are equidistant to M' upon light emission. M believes they are simultaneous Since they are equidistant to M' upon emission and they are simultaneous emissions in M, then they must also be simultaneous emissions in M' since there is no other basis to claim they are not simultaneous emissions for M' since they are equidistant at emission. Since they are equidistant and simultaneous in M', then they must reach M' simultaneously since the motion of the light sources has no bearing on the problem. But, they are equidistant and simultaneous in M also. Thus, both M and M' will be strick by the light beams simultaneously at different positions, a contradiction. Thus, either road you take you run into a contradiction.
swansont Posted January 29, 2010 Posted January 29, 2010 OK, you believe the satellites are not in sync from the view of the ground clocks. You believe also the emission are not in sync from M' and thus M' will believe B was emitted before A in order to preserve SR. This differential is caused by the distance to M' upon light emission. You believe they are not in sync based on distance to the clock. This is not supported by GPS but that is OK. Please don't strawman my position. The GPS analysis (that you claim is missing a time dilation term) is valid, and GPS works. That's my stance. If you read here, you will find the triangulation depends on the satellites have the same common time and earth time and then use distance to determine position. If the satellites were not in sync, this would not work. So, you are wrong in your thesis. http://www.beaglesoft.com/gpstechnology.htm Again, your summary of my position is incorrect, so this is pointless. Now, going back to the original R of S thought experiment from SR, both clocks are equidistant to M' upon light emission. M believes they are simultaneous Since they are equidistant to M' upon emission and they are simultaneous emissions in M, then they must also be simultaneous emissions in M' since there is no other basis to claim they are not simultaneous emissions for M' since they are equidistant at emission. For the last frakking time, NO. You CANNOT claim that they are simultaneous in M', you must SHOW that they are simultaneous in M'. And the math of relativity says that they are not! You have assumed the answer. That is an INVALID ANALYSIS. Since they are equidistant and simultaneous in M', then they must reach M' simultaneously since the motion of the light sources has no bearing on the problem. But, they are equidistant and simultaneous in M also. Thus, both M and M' will be strick by the light beams simultaneously at different positions, a contradiction. Thus, either road you take you run into a contradiction. The contradiction is in your assumption about simultaneity. It proves your assumption cannot be correct. Now, since your analysis is wrong, there is no point in applying it to any problem.
Cap'n Refsmmat Posted January 29, 2010 Posted January 29, 2010 (edited) I have a sneaking suspicion this thread is not going anywhere. Perhaps it would be best if I helped it. Watching this thread spiral toward its doom like a thirsty dog investigating a carnivorous toilet loses its fun after a while. vuquta, I'd hate to see another discussion end up in the toilet like this one has; in the future, please do read carefully what the physics experts say and consider that they may, in fact, be right. Thread closed. (flush) Edited January 29, 2010 by Cap'n Refsmmat
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