http://www.theflatearthsociety.org/forum/index.php?topic=62155.msg1702445#msg1702445
Citation provided.
I see an apparently empty claim from you about understanding frames of reference in that thread, but no actual evidence of same. Can you back your claim up?
While you're at it, I'm still waiting for your "correct" formula for velocity given acceleration and time. Can you provide it, please? You were the one who said my formula was wrong. Remember?
[Edit] typo.
V=AT simple unless you have a flat Earth a. That could do anything.
Actually, v = c tanh(aT/c)
Referring to the example in
http://www.theflatearthsociety.org/forum/index.php?topic=62155.msg1702445#msg1702445Q: what is the difference between v = aT and v = c tanh(aT/c) for a = 9.7755 m/s
2, t = 86400 s, and c = 299,792,458 m/s?
v = aT
= (9.7755 m/s
2)(86400 s)
= 844,603 m/s
v = c tanh(aT/c)
= (299,792,458 m/s) tanh(844,603 m/s / 299,792,458 m/s)
= (99,792,458 m/s) tanh(0.002817292)
= (99,792,458 m/s) (0.002817285)
= 844,601 m/s
A: It's down in the noise.
v = aT is perfectly adequate for the case given, especially considering the limited precision of the value for g
0 of 9.8 m/s
2 and variances of
about 0.5% from pole to equator in the example. We don't see any difference at all unless we carry a lot more digits than are justified by the input data. There's no need to use relativistic calculations for back-of-an-envelope examples like this. Sorry.