michel123456 Posted January 14, 2012 Share Posted January 14, 2012 You already made it: Since you have declared that: We know the speed at T1, T2 and T3, simply replace 1s with 1m/s, 2s with 2m/s and 3s with 3m/s. No. Or i don't get what you mean, or you don't get what I mean. You cannot simply replace Time by Speed. It is a Distance/Time diagram You cannot mix the one with the other. Your last post is very confusing. This diagram is a Distance/Velocity diagram (redshift=velocity as supposed). I understand that. I want to translate this into a Distance/Time diagram. Since Velocity is distance/time there must be no problem to do that. In a distance/time diagram, velocity is the tangent of the angle. Link to comment Share on other sites More sharing options...
Spyman Posted January 16, 2012 Share Posted January 16, 2012 No. Or i don't get what you mean, or you don't get what I mean. You cannot simply replace Time by Speed. It is a Distance/Time diagram You cannot mix the one with the other. Your last post is very confusing. Sorry if I was unclear Michel. You can use the data from a distance/time diagram to create a distance/velocity diagram. If you take the data from your distance/time diagram in post #72 and make a distance/velocity diagram, how does it look? I will repeat the data for you here: t=1s v=1m/s l=0.5m t=2s v=2m/s l=2.0m t=3s v=3m/s l=4.5m As you can see the values for time and speed are conveniently equal. Now please tell me if the data in your curve from post #72 have velocity proportional to distance? Link to comment Share on other sites More sharing options...
michel123456 Posted January 16, 2012 Share Posted January 16, 2012 (edited) Sorry if I was unclear Michel. You can use the data from a distance/time diagram to create a distance/velocity diagram. If you take the data from your distance/time diagram in post #72 and make a distance/velocity diagram, how does it look? I will repeat the data for you here: t=1s v=1m/s l=0.5m t=2s v=2m/s l=2.0m t=3s v=3m/s l=4.5m As you can see the values for time and speed are conveniently equal. Now please tell me if the data in your curve from post #72 have velocity proportional to distance? No it is not proportional. Here below the diagram of post #72 You are right, velocity is not proportional to distance in this diagram. A diagram where velocity is proportional to distance is (i hope being correct this time) the one from post#74 see below Which I don't fully understand, I have to admit. It looks like a locked situation. G1 cannot go to G2 unless at infinite speed. I don't know what this time 0,01389s represents. It is impossible to move G1 at another location at another time. I must be wrong somewhere. Edited January 16, 2012 by michel123456 Link to comment Share on other sites More sharing options...
Spyman Posted January 16, 2012 Share Posted January 16, 2012 I still don't understand what you are trying to do, but there is a difference if you have a time/distance diagram describing the movement of ONE object through several different phases or if you have a time/distance diagram describing ONE shot of several idividual objects in their respective different phase. Link to comment Share on other sites More sharing options...
michel123456 Posted January 16, 2012 Share Posted January 16, 2012 I still don't understand what you are trying to do, but there is a difference if you have a time/distance diagram describing the movement of ONE object through several different phases or if you have a time/distance diagram describing ONE shot of several idividual objects in their respective different phase. Yes. The last diagram is "ONE shot of several idividual objects in their respective different phase". It says nothing about precedent phases. Worse: it is locked. I wanted to do something and I got something else. Very accurately, I don't understand anything of my own diagram... Link to comment Share on other sites More sharing options...
michel123456 Posted January 20, 2012 Share Posted January 20, 2012 (edited) Aargh, I am 500 years too late.I rediscovered Galileo! look here "Galileo's rejection of the possibility of velocity changing uniformly with respect to distance" http://www.jstor.org/pss/226890 and http://books.google.gr/books?id=IgYADjGcJcAC&pg=PA130&lpg=PA130&dq=it+seems+in+his+eyes+plausible+that+a+falling+body&source=bl&ots=w9z1Zyw4hp&sig=itlLRv_Q4jz7RQ3mClGkoBLMhcU&hl=el&sa=X&ei=aLAZT4KEF8acOpm9jaAL&sqi=2&ved=0CCoQ6AEwAQ#v=onepage&q=it%20seems%20in%20his%20eyes%20plausible%20that%20a%20falling%20body&f=false --------------------- And the most interseting is this: If it is true that a single object cannot modify its state of motion in such a manner that velocity is proportional to distance, it means that Hubble's law is a demonstration that the objects were never in the same state of motion. Edited January 20, 2012 by michel123456 Link to comment Share on other sites More sharing options...
michel123456 Posted January 21, 2012 Share Posted January 21, 2012 (edited) Because I have a parallel conversation with someone else out of the Forum, who has deep objections. the basic equation of velocity proportional to distance is V=kD where V velocity, D distance, k constant of proportionality. And here below 3 diagrams for k=1, k=2, k=3 As you can notice, the points are always horizontal. The difference in time=zero ==> it is not possible for a single object to have such a state of motion: velocity proportional to distance is impossible for a single object. . Edited January 21, 2012 by michel123456 Link to comment Share on other sites More sharing options...
michel123456 Posted January 21, 2012 Share Posted January 21, 2012 (edited) edited. Edited January 21, 2012 by michel123456 Link to comment Share on other sites More sharing options...
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