1. Originally Posted by Aer
C does not "launch" D with any velocity other than .5c relative to A. C launches D with an acceleration equal to that of B. So u, should be calculated from:

.866c = (.5c + u) / (1 + .5u)

Whatever u is (I am lazy right now), it is not going to be .366c.
You are on the right track. However, you must not lose sight of the fact that "C" can become cloaked and vanish from the scenario such that it is clear that the physical flight schedule of D is that being side by side with B (which maintained constaned acceleration and increased 0.366c from the time D launched and remained side by side that D also increased 0.366c).

Uncloaking "C" they now want to claim D has a different rate of change. It does not and cannot not. Claiming the VAF is unjustified from any physics vantage point.

For D to have a terminal velocity of 0.866c using the VAF, u would have to = 0.6455c. But that is in consistant with the scenario stipulation that it launches at 0.5c when "C" went inertial wrt A.

2. Really, MacM, by continuing to insist that your u=0.366c figure is correct, you just make yourself look even more stupid.

3. MacM, B also would have the relative velocity of .6455c relative to C. Your analysis is wrong.

4. Originally Posted by MacM
Perhaps you should re-read more slowly or are you just going to ignore those that claimed 0.645c?
Why would I ignore them? They are correct.

Originally Posted by MacM
Oh, really. I think not bubba. I have not mixed frames. I have given a scenario and have discussed it from different frames of refereance and I am showing an inconsistancy in SRT.
Yes you have mixed frames, as Raphael pointed out.

Originally Posted by MacM
If you choose to deny that SRT requires any compound velocity "w" (i.e. - from A's view a missile (D) "velocity u" fired from a moving frame "C" "velocity v" has a terminal velocity of w = (v + u) / (1 + (vu /c<sup>2,/sup>) then have at it because you would look foolish indeed.
If you are taking v and U both from the same frame, which you are when you use .5c and .366c, then no you don't use the VAF. If you are taking v and u from different frames, in which case the values would be .5c and .646c, then yes you would use the VAF.

This really isn't that hard to understand, Mac. The VAF is for switching frames only. Using it for adding vAC and vCD from the same frame to find vAD in the same frame is simply a misapplication.

5. Originally Posted by MacM
Uncloaking "C" they now want to claim D has a different rate of change. It does not and cannot not. Claiming the VAF is unjustified from any physics vantage point.
It has nothing to do with "uncloaking", it has to do with changing from A's frame to C's frame. If you want to try to claim that all accelerations should be the same no matter what frame you choose to look from, go ahead, but you will be dismissed as the fool you are.

6. Originally Posted by James R
Really, MacM, by continuing to insist that your u=0.366c figure is correct, you just make yourself look even more stupid.
Really James R by continuing to insist that your u = 0.6455c figure is correct, you just make yourself look even more stupid. It is specified that C goes inertial at 0.5c and simultaneously launches D. D and B remain side by side and B increases (by physics) its velocity by 0.366c. They both have a common terminal velocity of 0.866c.

Hence by SRT D must have launched from C at 0.6455c not 0.5c. CONFLICT WITH REALITY, CONFLICT WITH REALITY ALARM. Not to mention the fact that C can cloak and vanish proving that B and D have identical flight schedules.

7. Originally Posted by Aer
MacM, B also would have the relative velocity of .6455c relative to C. Your analysis is wrong.
Correction SRT is wrong. You continue to recite SRT but provide absolutely NO physics justification for your claim. It was stipulated that D launched when C went inertial at 0.5c. For D to have a terminal velocty of 0.866c it would have to launch at 0.6455c from C. that is inconsistant with the scenario as stipulated. You are changing physics without justification.

8. Originally Posted by MacM
Really James R by continuing to insist that your u = 0.6455c figure is correct, you just make yourself look even more stupid. It is specified that C goes inertial at 0.5c and simultaneously launches D. D and B remain side by side and B increases (by physics) its velocity by 0.366c. They both have a common terminal velocity of 0.866c.
its called math, he did some (correctly by the way) to get the velocity WRT C. I got .6453, but it is probably rounding error in the difference there.

that why I asked you the question you dodged three times:
Originally Posted by cato
what velocity does D leave C at (WRT C)? you claim I am distorting, so straighten me out. answer?
it seems you are the only one having trouble doing math here.

9. For D to have a terminal velocity of 0.866c it would have to launch at 0.6455c from C. that is inconsistent with the scenario as stipulated

what is wrong with putting that velocity into your scenario?

10. Originally Posted by MacM
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You are changing physics without justification.
You are reciting rhetoric without any physics.

11. OK, Mac, answer this question then:

In C's frame, when C launches D, B has a velocity of 0 wrt to C, yes?

At the end of the scenario, what is B's velocity wrt to C in C's frame?

EDIT: please answer both what you think SRT predicts it will be and what you think it will be.

12. Originally Posted by GMontag
Why would I ignore them? They are correct.
Right. Now do the manly thing and justify your claim as I have repeatedly requested. I am very well aware of the claims of SRT, which is why I posted this scenario. But the issue is for you to justify your claim. Not merely parraot the claims of SRT.

I suspect you are incapable of doing that which is why you choose to start launching innuendos. Oh well. Others can see what is happening here and that is good.

Yes you have mixed frames, as Raphael pointed out.
Bullshit. I have specified the case from both frame views. I am demanding that you justify your claims of SRT.

If you are taking v and U both from the same frame, which you are when you use .5c and .366c, then no you don't use the VAF. If you are taking v and u from different frames, in which case the values would be .5c and .646c, then yes you would use the VAF.
You really are dense aren't you. Stop assuming you know so much more than other posters and pay attention. You are obligated to justify claiming the 0.645c figure in lieu of the 0.366c figure. I have shown that the flight schedules are the same for B and D. I have shown that B increases its velocity by 0.366c from the point of launch of D wrt A and frankly from C.

This really isn't that hard to understand, Mac.
You are right it is really quite simple. Only it is not justified and you have not justified it by physics.

The VAF is for switching frames only. Using it for adding vAC and vCD from the same frame to find vAD in the same frame is simply a misapplication.
Justify it.

13. Originally Posted by cato
its called math, he did some (correctly by the way) to get the velocity WRT C. I got .6453, but it is probably rounding error in the difference there.

that why I asked you the question you dodged three times:

it seems you are the only one having trouble doing math here.
Show any error in my math. The figure you quote is actually 0.6455c. The ones that seem to have trouble here are those that can't seem to justify in physics terms why one should accept 0.6455c as a velocity for D when it was launched at 0.5c and has a terminal veloicty of 0.866c.

Don't come back with the Frames BS. Justify the fact that B must increase 0.366c from the point that it was at 0.5c when C went inertial and that D was fired simultaneoulsy to that D is 0.6455c.

14. Originally Posted by cato

what is wrong with putting that velocity into your scenario?
Because it is inconsistant with the specifications of the scenario. To change it you must show just cause by physics. I have yet to see any one of you supply any justification what so ever, other than "Because that is what SRT requires". Hogwash.

15. Originally Posted by Aer
You are reciting rhetoric without any physics.
I suggest that claiming B increases 0.366c from the point it was at 0.5c physics not rhetoric.

I suggest that showing that D moves concurrent and side by side throughout the flight schedule is physics and not rhetoric.

I suggest that to claim that B and D have different velocities in this respect is rhetoric and not physics.

16. you ask us to justify, but you have not justified your claim, all you have done is some bad math and claimed it as good. you need to do some justification first.

I have shown that B increases its velocity by 0.366c from the point of launch of D wrt A and frankly from C.
what does "and frankly from C" mean?

17. MacM: "I suggest that claiming B increases 0.366c from the point it was at 0.5c physics not rhetoric." ... relativie to A, not C

MacM: "I suggest that showing that D moves concurrent and side by side throughout the flight schedule is physics and not rhetoric." ... You stipulated this, but yes, it is what happens in the scenario you provided.

MacM: "I suggest that to claim that B and D have different velocities in this respect is rhetoric and not physics." ... B and D have the same velocity wrt A, and that is .866c, they also have the same velocity wrt C, and that is .65c

18. Originally Posted by GMontag
OK, Mac, answer this question then:

In C's frame, when C launches D, B has a velocity of 0 wrt to C, yes?

At the end of the scenario, what is B's velocity wrt to C in C's frame?

EDIT: please answer both what you think SRT predicts it will be and what you think it will be.
By standard physics it has already been pointed out that you have a given mass, a given thrust and a given change in velocity 0.366c. Frankly which is relative to both A and C equally. It requires no different thrust or acceleration to achieve a 0.366c velocity wrt C than it does wrt A.

However, SRT would have you treat it differently when viewed from A if referance to C. If I claim v = 0.5c and u = 0.366c then SRT wants to claim w = 0.732c and we know that is nonsense since B actually has a velocity of 0.866c wrt A created by increasing velocity by 0.366c from the 0.5c point in the flight schedule.

w is velocity B wrt A, v is velocity of C wrt A and u is velocity of B wrt C.

19. Originally Posted by cato
what does "and frankly from C mean?
rhetoric

20. Originally Posted by cato
you ask us to justify, but you have not justified your claim, all you have done is some bad math and claimed it as good. you need to do some justification first.

what does "and frankly from C mean?
It means pragmaticly and in reality B's change in velocity wrt A and C is the same, inspite of the assinine claim of SRT.

BTW, your false assertions about bad math are just that false assertions. Now justify the physics of claiming the VAF.

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