I add 2 works more, probably more easy to understand:
4w - d/t=c in average.
We see objects at 13.7 billion light years, these objects are in distance at 13.7 billion light years and in time at 13.7 billion years, by that 13.7 distance/13.7 time = c = light speed = 1/1 . This give average of c.
According to theory double distance is double speed, but according by example:
1 hour - speed: 1 km/hour - distance 1 km
1/2 hour more - speed: 2 km/hour - distance 2 km - average is 2/1.5 = 1.33 km/hour // maximum is 2
1/2 hour more - speed: 4 km/hour - distance 4 km - average is 4/2 = 2 km/hour // maximum is 4
1/2 hour more - speed: 8 km/hour - distance 8 km - average is 8/2.5 = 3.2 km/hour // maximum is 8
None of theirs gives average of maximum speed
Only average speed is equal of maximum speed if the speed is constant.
By this expansion is impossible according to speed max= speed average and according to double distance is double speed
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5w - There is not form of double distance at double speed with speed incremented
According to the theory that says that at double distance is double speed and that speed is distance/time, at d/2 is c/2 and at d is c, how speed of d/2 is (d/2)/t, and d-(d/2) is d/2, need to expand at half time in the form (d/2)/(t/2), this is c, but c is average and not maximum, by that only at constant speed is possible.
In incremented is impossible because an average from 2 is by example 1.5 to 2.5, but 2.5 is not double that 1.
An example:
1 hour accelerating from 10 to 20 km/hour - speed final 20 km/hour - distance 15 km
1 hour accelerating from 20 to 40 km/hour - speed final 40 km/hour - distance 45 (30+15)
Really the relation is 4d is double speed
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Also I arrange 1w:
1w – The light arrive to us in less time that need the expansion theory.
Resume: Expansion is from d/4 to d/2 (for example) with different speed (in d/4 expansion is half of d/2 according to the theory that relation distance and speed) and against this the lightspeed is always the same (constant speed): by this the light from d/4 arrive to us before to stay in d/2.
At double distance is double speed, according to the theory.
An example with d in 16 = c (lightspeed) and by that d/2 is 8
Light from d/2 need t/2 time to arrive to us, in the example d/4 (example 4) in t/2 is d/2, lightspeed is 16.
t/2 is 2 times t/4 - in t/4 light travel 4
first t/4 - light travel 4, distance with expansion is 3+
Light of d/4 in t/2 is d/2, in d/4 speed is (d/4)/t and in d/2 need to be double speed - this is another (d/4) in half time t/2 so (d/4)/t is half speed that (d/4)/(t/2) (increment is d/4
So a photon emitted in d/4 made t/2 time arrive to us t/2 time from d/2 according to the theory : double distance is double speed and d/2 is a c/2.
I divide in the example the time in t/8 so expand 1 and light travel 2 (according to the theory we can consider that in any t/8 expand 1, [really the first would be less because it's half speed], so 4t/8=t/2)
1º t/8 : 4 +1(expand) - 2 (light travel) = 3 distance (4 expand 1 in t/8)
2º t/8: 3 + 0.75 (expand) - 2 (light travel) = 1.75 - (here really would expand 1 in relation to 5, not 4)
3ª t/8 : 1.75+ 0.44 (1.75/4) - 2 (light travel) = 0.19 - (here really would expand 1 in relation to 6, not 4)
So light arrive in few more that 3t/8 and according to the theory arrive in 4t/8 = t/2
In this example not consider that a part of travel of light is not expanded or stretch, also that this light is near us and by that probably there is less stretch, ... and by that less time
A same photon cannot arrive in 2 different moments, by that the expansion theory is not possible.
Thanks and sorry by disturb you.
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