Forums before death by AOL, social media and spammers... "We can't have nice things"
|    sci.physics.research    |    Current physics research. (Moderated)    |    17,516 messages    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
|    Message 15,865 of 17,516    |
|    Tom Roberts to John Heath    |
|    Re: the limits c    |
|    30 Sep 17 11:10:00    |
      From: tjroberts137@sbcglobal.net              On 9/29/17 9/29/17 8:20 AM, John Heath wrote:       > A car is moving at .1 c.              RELATIVE TO THE ROAD. When you don't specify which frame you are using you       only confuse yourself (and any readers). Here we presume the road is at rest in       an inertial frame.               Compared to a car at 0.1 c, the earth's rotation and orbital        speed are negligible. Not to mention that it would launch into        space. So assume a flat road sufficiently long.              > The tire on the road at 6:00 o'clock , bottom , is moving at 0 c as it is       > touching the road but at 12:00 o'clock , top , the tire is moving at .2 c       > twice as fast as the car. Does this mean the car is limited to speed .5 c as       > at this speed the top of the tire will be moving at c twice as fast as the       > car ?              No. You have assumed Galilean relativity, in which velocities add vectorially.       That does not correspond to the world we inhabit.              You must use the Lorentz composition of velocities: Both points of the tire are       moving at 0.1 c relative to the axle (in opposite directions), and the axle is       moving at 0.1 c relative to the road (along with the rest of the car except its       wheels/tires). So the top of the tire is moving at this speed relative to the       road:        v = (v1 + v2) / (1 + v1 * v2 / c^2)        v = (0.1 c + 0.1 c) / (1 + 0.1 * 0.1) = 0.198 c              If the car moves at 0.5 c relative to the road, the top of the tire moves at       this speed relative to the road:        v = (0.5 c + 0.5 c) / (1 + 0.5 * 0.5) = 0.8 c              If the car moves at 0.99 c relative to the road, the top of the tire moves at       this speed relative to the road:        v = (0.99 c + 0.99 c) / (1 + 0.99 * 0.99) = 0.9999 c              For all car speeds relative to the road that are less than c, the top of the       tire also moves relative to the road at a speed less than c.               Note I have ignored the impossibility of such speeds using tires        -- no material is anywhere close to strong enough to sustain        the strain induced by such enormous rotational velocities.              Tom Roberts              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
(c) 1994, bbs@darkrealms.ca