[alancalverd]

"I drove a lap of a circuit in my Ford Fiesta at 60 mph, how fast do I have to go in the second lap to double my speed?"

Second lap speed = x.

Average speed = (60 + x)/2 = 120 

x = 180.  So I wouldn't use a Fiesta, but no problem with a F1 car or a small plane.

I suspect you have not formulated the question quite as you intended.

[Halc]

"I drove a lap of a circuit in my Ford Fiesta at 60 mph, how fast do I have to go in the second lap to double my speed?"

Second lap speed = x.

Average speed = (60 + x)/2 = 120 

x = 180.

The Fiesta then completes the first lap in time t and the second lap in t/3 for a total of 4/3t for 2 laps, or 2/3 t average per lap, when you'd need 1/2 t average to have the speed double.

You are one of many who don't know how to compute average speed. Average speed is total distance/total time.
You are spending only a fourth of your time going the fast speed, and yet you are averaging it equally with the time spent going 60, a classic mistake.

[Bored chemist]

"I drove a lap of a circuit in my Ford Fiesta at 60 mph, how fast do I have to go in the second lap to double my speed?"

I did something.
How fast do I have to do it to double my speed?

Well, obviously, twice as fast. That's what doubling your speed means.

The "puzzle" doesn't say anything about averages, does it?

You are one of many who don't know how to compute average speed.

Nobody was asked to.
So Alan's comment, and yours don't make sense.

I think this

I suspect you have not formulated the question quite as you intended.

may be the salient point.

[Bored chemist]

Construct a device in the shape of an equilateral triangle. Place two lasers at corner A and targets at corners B and C. Have a second laser at corner B and a clock at corner C. Fire the two lasers at corner A simultaneously. One at corner B and the other at corner C. When the light is detected at corner C start the clock. When the light is detected at corner B fire the laser at corner B towards corner C. When the light from corner B is detected at corner C stop the clock. This device will work in any orientation to measure any discrepancy in the expected value for the speed of light.

You might want to draw a diagram.
It won't stop you being wrong, but it might help us explain why you are.

[alancalverd]

You are one of many who don't know how to compute average speed. Average speed is total distance/total time.

No, that's speed averaged over a fixed distance. Average speed is the average of all speeds.

[Eternal Student]

Construct a device in the shape of an equilateral triangle......

   I loved that as an idea and it seemed for a while that it could even work.  Well done jeffreyH.

This seems to be the diagram as requested by Bored_Chemist:

      A - - -  - - - B
        \             /
         \          /
           \      /
              C

But equilateral and done with a graphics package that I can't be bothered with.  Twin lasers at A and clock and C.

The problem seems to involve firing the twin lasers from A.    You can't assume that  light has reached all the way to B  just because it has reached C and caused the clock to start.   It may be that the light hasn't reached B yet (if light travels slower in the AB direction)  or alternatively that B has already fired its laser and light is half-way up the line BC before the clock starts (if light travelled faster along the AB direction).
   So it seems that what you're measuring with the time interval on the clock at C isn't how long it takes light to travel along BC but instead how long it takes light to travel an unknown distance and along what could be two different directions.
   You'd have to write out the maths and tell us more about exactly what you were hoping to measure,  JeffreyH,  before I can be certain what you had hoped to do with that apparatus.

[jeffreyH (OP)]

But the distance BC is known and we know the speed of light. Therefore, we also know what time the clock should show if light travels each path at the same rate. If it doesn't read the time we expect then something is amiss.

[jeffreyH (OP)]

BTW It might show the correct time in every orientation, which would also be interesting.

[Bored chemist]

we know the speed of light.

No we don't.

[jeffreyH (OP)]

we know the speed of light.

No we don't.

OK but now you are being pedantic. We know what the speed of light in a vacuum should be.

[David Cooper]

It should be fairly easy to calculate the one-way speed of light.  The problem is doing so with any reasonable amount of accuracy.

Here's how I would do it.


SpeedOfLight.gif (9.32 kB . 669x311 - viewed 46915 times)

Light (strong laser) passes through two shaft connected spinning discs, hits a cylindrical (conical) mirror, and is projected onto a wall to record.  Mostly interested in the trailing edge of the light spot.

The shaft will twist if it's aligned with its direction of travel, so the slits won't both be at the top or bottom at the same time as each other. If you imagine two clocks at either end of the shaft with their dials aligned with the discs and with a nanosecond hand going round, the alignment of the apparatus and its speed of travel will affect the synchronisation of the two clocks, so the hands, just like the slits in the discs, will not always both be at the top or bottom at the same time, even if they are when the apparatus is at rest. Move it to the right and the clock (and disc) to the right will lag in its timing compared to the clock (and disc) to the left. This, in combination with length contraction, will always completely mask the differences you're trying to measure.

[yor_on]

Thanks for the link wolfekeeper. It was fun. you could also assume that 'c' is local, meaning that it is related to your local frame of reference even in a two way mirror experiment, probably much the same as he was discussing in that movie. And if you then equal that to a perfect local clock, aka 'split' 'c' as one, then no clocks can be said to be the same, unless they have a exact same ideal frame of reference. And that ideal frame must then become extremely 'local' when I think of it.  'Quantum local' :)

And then you have that possibility in where it all becomes a sham. No 'propagation' at all, just a ideal expression, to us defined as a 'speed' inside our four dimensional SpaceTime.


spelling
=

Thinking of it, you might even be able to join those two into one. As that 'quantum dot' representing a ideal local 'tick' or 'clock' should be equal to all other, locally defined, 'ticks' or clocks. Assuming it's scalable into a macroscopic universe. you would then have a 'absolute time' scaling down, and a 'relative time' scaling up. Crazy, isn't it :)

[Halc]

David,
Excellent analysis of the rotating cylinder thingy. Yes, per RoS, a rigid rotating cylinder will be twisted relative to a frame in which it has linear motion along its axis of rotation, which is not immediately intuitive.

Meanwhile, the line of clocks to measure expansion thing contradicts established science and has been split off here:
https://www.thenakedscientists.com/forum/index.php?topic=82441.0

[David Cooper]

In moving all that, you took this with it, so I'll put it back here:-

Construct a device in the shape of an equilateral triangle. Place two lasers at corner A and targets at corners B and C. Have a second laser at corner B and a clock at corner C. Fire the two lasers at corner A simultaneously. One at corner B and the other at corner C. When the light is detected at corner C start the clock. When the light is detected at corner B fire the laser at corner B towards corner C. When the light from corner B is detected at corner C stop the clock. This device will work in any orientation to measure any discrepancy in the expected value for the speed of light.

Let me add a point D half way along the line BC. I'm doing this so that I can describe the direction AD. If the apparatus is stationary or moving in either direction along the line AD, the light from A will reach C to B simultaneously. If they both send a pulse at each other, they will meet at D while passing each other and reach their target corner simultaneously too.

What happens though if the apparatus is moving in the direction BC instead? Well, the light from A will reach C before it reaches B, so C will start its clock before B sends out its signal and, when you combine this with the length contraction acting on the equilateral triangle (which is now contracted in the BC direction), the timing difference that you're trying to detect will be completely masked.

All experiments of this kind are rendered useless by the masking that is bound up in the phenomenon of relativity, apart from the type that's been described at the link in reply #52 where established science conflicts with itself.

[Just thinking]

one way speed of light ..PNG (34.18 kB . 896x849 - viewed 4088 times)Yes by using the earth and the moon to view a distant moving object. If we were to place a telescope with a camera on the moon with a transmitting device that can detect and tracked Jupiter and register the very moment that one of the moons of Jupiter has just alined its edge with the edge of Jupiter sending a registration tone back to earth at that very moment we would no the one way speed of light. How, well here on earth using simple mathematics we would no the very moment that this alignment would take place on the moon without being there. So we could say the alignment is taking place now and see if the message from the transmitter arrives at the expected time. The time would be expected to be according to the distance of our moon from the earth at that moment in time. The experiment could be carried out by a maned or unmanned mission.

[Just thinking]

No we don't.

Can I challenge you Bored chemist to see if you are able to pick apart my method of measuring the one way speed of light? Good luck.

[Kris Kuitkowski]

I think we can measure one way speed of light.
we need to redefine simultaneity. My proposition is as follows:
if a rigid body AB of length l perpendicular to X axis is moving without any acceleration parallel to X axis and at time t0 its point A is at location (x,0) then simultaneously its point B is at location (x,x+l)

Let's now design the experiment to synchronize distant clocks and measure one way speed of light:

Imagine four spaceships flying as perfect square EFGH towards (or away from) not moving (at least relative to each other) points ABCD, where AD is parallel to EF and distance EF equals AD. Points EG should be collinear with points AB and points FH collinear with CD. Making sure that ABCD (and EFGH) is a square is relatively easy, since 2-way speed of light is constant:
we can measure (and correct, if necessary) distances BD and CA by sending light signals from B to D (and from C to A) and back
Now at certain time (clocks at A and D can be pre-synchronized using Einstein convention, but it is not absolutely necessary) we can measure distance from A to G (L) and from D to H (L’) using light (laser) signal send from A to G (and reflected back to A) as well as distance from D to H. If the distance AG (L) equals DH (L’) signals from A and D had been sent simultaneously; if not, it would be easy to adjust the clocks so they are synchronized.
Please let me know if I made any wrong assumption

Of course, theoretically it would be sufficient to have only the lines AD parallel to GH, but practically it could be difficult to make sure they are parallel to each other

[Bored chemist]

Can I challenge you Bored chemist to see if you are able to pick apart my method of measuring the one way speed of light?

Are you sure you want to?

[Eternal Student]

Hi Kris, I hope you are well.

we need to redefine simultaneity. My proposition is as follows:
if a rigid body AB of length l

   I'll re-read your article later when I have more time.  Straight off the bat my first concern is that you are using Rigid Bodies in the context of Special Relativity.   They just don't behave (or exist) in the way you may have hoped.

See,  for example:   https://einstein.stanford.edu/content/relativity/q2018.html   for a short explanation of the problem.
Or the Wikipedia entry about  "Born Rigidity", although this is a bit more involved or complicated.

None the less,  you've got some good ideas and thank you for spending some time here.  It doesn't seem as if you are asking the rigid bodies to accelerate so you may not run into too many problems, I'll have to re-read what you've written later.

   Best wishes to you.

LATE UPDATE:   I've had a bit more time to read and look at the diagram.    The distance L'  does not seem to be  the distance  DH as  you have written but instead it is the distance DH'  .   I can't follow the diagram, sorry.

[Just thinking]

Quote from: Just thinking on Today at 15:26:08

    Can I challenge you Bored chemist to see if you are able to pick apart my method of measuring the one way speed of light?

Are you sure you want to?

Yes please it would be all my pleasure.