# light propagates at c + v?

Carpenter maths and elementary school logic is all thats needed to understand mine.

Prove why the local observer and remote observer see different things and I will believe you, but you going to need solid logical physical evidence.

Relativity is easy to understand, I just dont agree with it.

Your Italic cryptic lingo isn't helping you either. What you going to do now?
Have you achieved extremely high speeds recorded the data and proved your assumptions? Till then all relativity has is pseudoscience, yes? just like curved spacetime and the rest of it all pseudoscience, yes? Mods better move all the posts in the science section to the fringe sections. Mainstream deals with "special" fringe science... lol omg!
Im showing you at low speeds the claimed effects are non-existent, why should they manifest at high speeds if no trace of them exists at low speeds?

if you want to achieve extremely high speeds ask me how and if I feel motivated I might just tell you
At first it looked like you were up-to-speed on the established body of work, and were simply finding a discrepancy in one aspect.
But you've tipped your hand, and admitted that your elementary school education is preventing you from exploring further than a Newtonian classical world.
Instead of learning, you simply deny everything you don't understand, down to and including relative velocity.

We can help people who are willing to learn, but we cannot do anything about the willfully ignorant.

This has stopped being a discussion.

The faster traveling satellites clock measures a change in time because at a faster velocity the frequency doesn't change the distance between the source and emitter has changed via the velocity (or real time motion) or in other words it has a longer/shorter path to travel and you think its a change in frequency. in other words the distance between the source and emitter has a longer/shorter path to travel... Which is not a change in time at all...
For this now I need to calculate the amount of time dilation the satellite clock time will dilate compared to the earth sea level clock in this theory and see if it matches current values of compensation/dilation.

Because the one beam has a shorter path to travel than the other beam its in realtime motion.
While the one beam going in the direction of travel has a longer path 1m + 0.767188mm to travel , the beam going in opposite a shorter path 1m - 0.767188mm

Photon a travels 1m + 0.000767188m = 1.000767188m
Photon b travels 1m - 0.000767188m = 0.999232812m
Photons a speed is c + 230000m = 300022458m
Photons b speed is c - 230000m = 299562458m

Photon a travels the distance 1.000767188m / 300022458m = 3.3356409 20587351e-9ns
Photon b travels the distance 0.999232812m / 299562458m = 3.3356409 83423898e-9ns

thats what I get when calculated using the values from the OP.

Lets calculate at only c then, as you suggest is correct.

Photon a travels 1m + 0.000767188m = 1.000767188m
Photon b travels 1m - 0.000767188m = 0.999232812m
Photons a speed is c 299792458m
Photons b speed is c 299792458m

Photon a travels the distance 1.000767188m / 299792458 = 3.338200015692189e-9ns
Photon b travels the distance 0.999232812m / 299792458 = 3.333081888270852e-9ns

Which theory shows a greater drift the c or the c + v?

And you going to say the unrealistic scenario of the following is correct, because its the only thing you can say in your defense. That the path length of the photon to the local observers frame doesn't change, but it changes for the remote observer.

Photon a travels 1m
Photon b travels 1m
Photons a speed is c 299792458m
Photons b speed is c 299792458m

Photon a travels the distance 1 / 299792458 = 3.33564095198152e-9ns
Photon b travels the distance 1 / 299792458 = 3.33564095198152e-9ns

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Instead of learning, you simply deny everything you don't understand, down to and including relative velocity.

We can help people who are willing to learn, but we cannot do anything about the willfully ignorant.

This has stopped being a discussion.
That seems to be a common theme among our cranks, although in some, more common than others.

And you going to say the unrealistic scenario of the following is correct, because its the only thing you can say in your defense. That the path length of the photon to the local observers frame doesn't change, but it changes for the remote observer.

So in GR the local observer is going to see this.
Photon a travels 1m
Photon b travels 1m
Photons a speed is c 299792458m
Photons b speed is c 299792458m
Photon a travels the distance 1 / 299792458 = 3.33564095198152e-9ns
Photon b travels the distance 1 / 299792458 = 3.33564095198152e-9ns

And the GR remote observer is going to see this.
Photon a travels 1m + 0.000767188m = 1.000767188m
Photon b travels 1m - 0.000767188m = 0.999232812m
Photons a speed is c 299792458m
Photons b speed is c 299792458m
Photon a travels the distance 1.000767188m / 299792458 = 3.338200015692189e-9ns
Photon b travels the distance 0.999232812m / 299792458 = 3.333081888270852e-9ns

In GR the time changed because the path length increase for the remote observer only and it says their clock runs faster/slower.

I dont think so, but let me explain why the faster clock records a different time.
We calibrate the clocks at sea level and the one leaves on the eg.satellite and starts traveling at a greater velocity than the one left on Earth. The greater velocity of the satellite clock causes the path length of the satellite clock photons to change and the detector records the frequency changing. The recorded time between the two clocks desynchronises.

The problem I find in GR is it assumes the remote and local observers see different things. To the c + v theory they see exactly the same thing only the detector on the faster moving device takes longer to record the frequency because of the increased path length due to greater velocity.

I cant find the data I need to calculate the real satellite dilation.

In other words both observers see this for earths (local) clock at c + v
Photon a travels 1m + 0.000767188m = 1.000767188m
Photon b travels 1m - 0.000767188m = 0.999232812m
Photons a speed is c + 230000m = 300022458m
Photons b speed is c - 230000m = 299562458m
Photon a travels the distance 1.000767188m / 300022458m = 3.335640920587351e-9ns
Photon b travels the distance 0.999232812m / 299562458m = 3.335640983423898e-9ns

And both observers see this for the faster (remote) clock at c + v. This is an example and not real values...
Increasing values buy 20% for the example.
Photon a travels 1m + 0.0009206256m = 1.0009206256m
Photon b travels 1m - 0.0009206256m = 0.9990793744m
Photons a speed is c + 276000m = 300068458m
Photons b speed is c - 276000m = 299516458m
Photon a travels the distance 1.0009206256m / 300068458m = 3.335640914314293e-9ns
Photon b travels the distance 0.9990793744m / 299516458m = 3.335640989718168e-9ns

The difference in the counted time between the Earth at 230000m/s and faster remote clock at 276000m/s would be
1 sec on earth is 0.999847 sec on the faster moving clock.
Is that close to relativity predicted values?

The difference in the counted time between the Earth at 230000m/s and faster remote clock at 276000m/s would be
1 sec on earth is 0.999847 sec on the faster moving clock.
Is that close to relativity predicted values?
Where have you stated the speed of the satellite? Without knowing that, one cannot determine the dilation factor.

The factor you list correlates with a relative speed of 3250 miles per second, which is about 420 times too fast for an orbiting satellite.

Note that you are only considering SR, and not taking GR into account. GR is actually a laarger contributing factor to time dilation than SR.

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So in GR the local observer is going to see this.
Photon a travels 1m
Photon b travels 1m
Photons a speed is c 299792458m
Photons b speed is c 299792458m
Photon a travels the distance 1 / 299792458 = 3.33564095198152e-9ns
Photon b travels the distance 1 / 299792458 = 3.33564095198152e-9ns
This is correct.
And the GR remote observer is going to see this.
Photon a travels 1m + 0.000767188m = 1.000767188m
Photon b travels 1m - 0.000767188m = 0.999232812m
No that is incorrect. The remote viewer would say that both photon a and b traveled 1meter.
Photons a speed is c 299792458m
Photons b speed is c 299792458m
Right in m/sec
Photon a travels the distance 1.000767188m / 299792458 = 3.338200015692189e-9ns
Photon b travels the distance 0.999232812m / 299792458 = 3.333081888270852e-9ns
Since we now know they traveled the same distance this question is moot.

No that is incorrect. The remote viewer would say that both photon a and b traveled 1meter.
Just to be clear both photons would move 1 meter in 3.33564095198152e-9ns by the remote viewers clock because both photons have a speed of c.

No that is incorrect. The remote viewer would say that both photon a and b traveled 1meter.
That is for the local clock to complete the distance of 1 cycle for the remote observer. The speed is set to c = 299792458.

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Just to be clear both photons would move 1 meter in 3.33564095198152e-9ns by the remote viewers clock because both photons have a speed of c.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift. The only time you are ever going to see this \ is if the local observer is constantly accelerating when the photon leaves the detector. Which proves that when a photon is emitted it is no longer associated to the local frame of reference, because when the photon is in on route and we accelerate the photon doesn't inherit the new velocity.

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That is for the local clock to complete the distance of 1 cycle for the remote observer. The speed is set to c = 299792458.
What?

if photons only moved at c then the local and remote observer would see this \ instead of this | and this /,
Not completely sure I get your way of explaining this. But I think using you method this is how to explain it. The local observer sees this | and the remote viewer sees this /. By remote viewer I mean someone who is in a different inertial frame.

... if the clock had great enough velocity and we would detect a drift.
FOR THE LOVE OF ALL THAT IS HOLY - FALSE.

!&@#&* 271 posts of this \$!&#\$!*@

The only time you are ever going to see this \ is if the local observer is constantly accelerating when the photon leaves the detector.
Source
What what?
Not completely sure I get your way of explaining this. But I think using you method this is how to explain it. The local observer sees this | and the remote viewer sees this /. By remote viewer I mean someone who is in a different inertial frame.
You catch on quick, wow, Im so proud of you...
FOR THE LOVE OF ALL THAT IS HOLY - FALSE.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
Where have you stated the speed of the satellite?
!&@#&* 271 posts of this !&#!&#!*@
(c-v)+v ?

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What what?
This is an unclear statement:
That is for the local clock to complete the distance of 1 cycle for the remote observer. The speed is set to c = 299792458.

You catch on quick, wow, Im so proud of you...
No need to be a dick.

if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.
Each one of those scenarios is wrong. The local observer sees this |and the remote observer sees this /. That of course assumes the local frame is moving to the right relative to the remote frame.

You are stuck in a 19th century classical Newtonian universe, where space and time are absolutes.
All of your assumptions are more than a century out-of-date.

if photons only moved at c then the local and remote observer would see this \ instead of this | and this /, and the photon would miss the detector if the clock had great enough velocity and we would detect a drift.