[guest39538]

@David

Why do we need any maths when we can have a scale system?  2:1 ratio

c-graph.jpg (78.8 kB . 1445x505 - viewed 4841 times)

Does your maths agree with my diagram? for the first part which is front to rear.

Your first answer must be around the 1.7s mark. If not you got it wrong somewhere.

added- huh, somethings not right, how can they meet in 1.7s when the light has travelled for more than that , scratches head and rethinks diagram.

I might have to agree time slows down .

Correction , the first contact point should be on the 2 second mark.

p.s damn, i messed it up lol, i will do it again later,

[Le Repteux]

Hi guys, hi David, 

I'm Raymond, the guy on Facebook. :0) Your nice simulation on Magic Schoolbook shows very clearly how and why length contraction should occur, so I take it as a possibility, but you also agree with relativists that a light clock should slow down when it moves since light takes more time between the mirrors this way, and I can't see how it should since there would be no doppler effect to measure whatever the speed, thus no variation in the elapsed time between the light pulses. You say «The simplest kind of clock is the light clock: this is a device which sends a pulse of light out to a mirror which then reflects it straight back again to a sensor from where the light was originally emitted, and the round trip of the light pulse counts as a tick. ». But it seems to me that, without doppler effect, even if light would take more time between the mirrors, the distance between the tics would always be measured the same whatever the speed, which means that a clock that has traveled during a certain time would show exactly the same elapsed time than one at rest.

[guest39538]

ruler.jpg (96.98 kB . 1765x505 - viewed 4806 times)

added- still not quite right.

tA=0.75s

tB=1.5s

thinks that's correct now?

Anyway David, what i have been trying to explain to you is that it means nothing.  There is no contraction of space or the carriage. There is only objectively  a variance in distance that gives you a variance in tick rate.

[David Cooper]

Quite clearly you think a clock is more than it is.   I know what a light clock is but quite clearly you ignore the observer affect by adding mirrors and such, setting the parameters to fit the ''story'' without considering the what at best you have in terms of objective reality.  The clock does not tick slower to begin with, we can work out the extra distance the light needs to travel and adjust accordingly to maintain the same rate of tick. i.e 1 second would not be equal to 1 second unless we calculated the difference to synchronise the difference.
I.e one clock would be measuring 1 second while 1 clock was measuring 1.2 seconds. The duration and length of a second then remaining the same with no contraction needed .

Picture this. We have two light clocks sitting next to each other, and let's align them the same way as each other. We move one of the light clocks away from the other and it ticks less often than the stationary one because of the increased distance light has to travel in the moving clock to complete each tick. We stop moving that clock and see it return to ticking at the same rate as the clock that never moved. Next, we move it back again to reunite it with the stationary clock, and while it's moving back it ticks more slowly again. Once we have brought it to a halt next to the clock that never moved, the two clocks tick in sync with each other again. If each clock has a counter, as clocks usually do, they might both have been set to zero at the start of the experiment. Now one of the clocks has a higher reading on it than the other because it has recorded more ticks. The clock that did all the moving has registered fewer ticks. The light in both clocks travelled exactly the same distance through space during the experiment, but it had to go further in the moving clock for many of the ticks and therefore couldn't complete as many ticks for its clock as the light in the stationary clock.

To claim that the clock that moved didn't tick more slowly than the one that stayed still the whole time is a nonsense - the different counts that they've racked up prove it.

[David Cooper]

You tell me 0.867 rounded off 0.866025. So rather than 1.33 that you gave me you used the correct 1.33975 for 7.461. We are just using ratio's for distance light travels. How we measure time will fall out of our measurements as ratios. The end result at this speed will be half the tick rate. You are not following the postulates of relativity in your understanding for distance. Your using a fudge factor and confusing yourself with your current understanding of time.

The same distance as what? The time that you mistook for distance (and whose value is not quite 7.5)? You've made a massive error which you're now trying to build upon.

Rounding off is a massive error? I see you rounded up to 0.867 when .866 was more accurate

You're right about my 0.867 being wrong - I've accidentally got into the habit of using the 7 from the 0.87 version. Every time I use the number in a calculator though, I use sin(60) and never type in a rounded off version of it, and I'd expect you to do the same. The important point here though is that the 7.461 figure is not 7.5, but you appeared to be mistaking your rounded up 7.5 for sin(60)^2. I was drawing attention to the difference in order to help you see that it is not what you thought it was, the 7.461 actually being the time of the forward trip of the light. I have used no fudge factor in my calculations and am not confused in the least - you imagine that the massive error I referred to is the rounding error, but that was just a small detail. The massive error was you taking 7.5 (or the correct value of 7.461) as the distance the light has to travel from the rear mirror to the front one - the distance it actually has to travel is TEN TIMES as great: 74.61cm. Look at the first of my interactive diagrams and watch how long it takes for the red pulse of light to travel from the rear mirror to the front one and see how far across the screen it has to travel during that time.

For the return trip, the carriage moves 53.5898 and the light moves 4.641cm before they meet. I can't make sense of what you're trying to do there with any of what you've done there.

I don't know how, but I made two mistakes in that bit. If you knew the right numbers though, you should have been able to work that out yourself: the decimal point is in the wrong place for the first number, and the two numbers need to be switched over. The carriage moves 4.641cm and the light moves 5.35898. And before you try to capitalise on that and say it's as bad as your error, it isn't - I didn't build anything upon these errors and it's only the corrected versions that fit with what I've been saying, but you built a lot on top of yours and correcting them will force you to throw out your argument.

Of course you cannot make sense of what I am saying. You do not understand the ratio for closing distances. You are probably subtracting your physical contraction but here is the real closing distances.

There is no physical contraction involved in these calculations - we haven't got that far yet. These calculations are all about an uncontracted train.

Follow my logic: if one side is light speed and the other side is light speed they would meet in the middle 50% or .50 / .50 for distance covered.

You've got that bit right.

Now if the mirror on the physical object is moving at 0.866025c and closing on c, c wins. The physical object moves 0.866025 for every 1.0 for light. So the closing speed of light has to be over 50% of the closing distance. I just estimated 57% which was 0.57 of one length as a ratio. Light always wins in closing ratios.

I had assumed you had calculated your 57% figure rather than just guessing, and that's why I thought you had applied an incorrect method. I gave you my numbers (one incorrect, and both incorrectly labelled) in order to give you something to aim for in your calculations, and if you'd done the maths properly you would have understood where my numbers come from despite the errors, and you would have known that we had both applied a valid method.

Now light goes forward 7.46 and returns 1 for light and 0.866 for the object (clock) we have 0.133 difference divide that by 2 for a quick estimate gives about 0.066 which I rounded up to 0.57 and added it to my rounded off 7.5 cars for light to catch the front mirror for light to travel 7.46 + 0.57 = 8.03 for its length. Now we divide it by 2 for 4.015 cars divide into one car and we get 0.25 the ratio. The square rt. of 0.25 is 0.5 tick rate.

That is a hopeless mess of misapplied maths! The correct method is to add the speeds of the train and the light [or subtract one from the other if you're taking one of them to be negative] which gives you a closing speed of 1.866c. You then divide the separation distance (10cm) by that speed to get the time taken for the rear of the carriage and the light to meet up. You then take that result and multiply it by c (=1) to get the distance travelled by the light in that time. To get the distance travelled by the carriage in that time, you can either multiply the same result by 0.866 or you can take away the distance travelled by the light in that time from the 10cm initial separation figure - both methods produce the same value.

If you are unhappy with Lorentz I am unhappy with physical contraction. You do not fudge objects to fit math when you do not understand what time represents

I am very happy with Lorentz, and we haven't done any physical contraction here yet. You are the one doing the fudging with your crazy misapplication of maths where you've defecated a whole lot of numbers into a pot and stirred them together with a determination to make them conform to the right amount of time dilation for a perpendicular clock which is the wrong answer for this uncontracted clock aligned lengthways with the train's direction of travel. I've rarely seen maths being so misused outside of politics. It's shocking, and it will do your reputation here no good at all if you fail to correct it in a hurry. Do the job properly!

Not possible - light takes 2t for the round trip on both clocks with the carriage at rest. With the carriage moving at 0.867c, it takes 4t on the perpendicular clock, and 2t for each half of that, so it reaches the far perpendicular mirror in 2t and reaches the front mirror of the other clock in 7.4641t.

Just like you missed the closing speeds your missing the position of the perpendicular mirror in the clock at 0.866025c when the light reaches the mirror. The perpendicular mirror reaches the position of 7.46 cars the light still has not reached the perpendicular mirror. The perpendicular light only found space and not the mirror when light reached the mirrors position from the past. The angle of light is still traveling to hit the angled closing position in space.
The photon has to follow the hypotenuse and has not reached the opposing mirror by the 7.5 forward ratio.

You still haven't taken in the scale of your monumental error! The 7.4641t is a time and not a distance. The distance is that time multiplied by 10 (which comes from the 10cm distance). The train moves over seventy four cm before the light reaches the front mirror.

It reaches the mirror long before the 74.641cm point which is the distance you should be using.

Your logic is missing the mark.

Your maths is woeful, and your eyesight may be just as bad. Look at my interactive diagrams properly and measure the distance the red dot moves across the screen from the point where it leaves the rear mirror to the point where it catches the front mirror. Then compare that with the distance the apparatus moves between the other red dot leaving the bottom mirror and arriving at the top one. Maybe your screen isn't wide enough to do the first measurement? It that's the problem, just use the lower interactive diagram that shows the length contraction applied to the MMX, then double the distance that you measure there (but don't double the other distance).

[David Cooper]

Hi Raymond,

...but you also agree with relativists that a light clock should slow down when it moves since light takes more time between the mirrors this way, and I can't see how it should since there would be no doppler effect to measure whatever the speed, thus no variation in the elapsed time between the light pulses.

I've answered part of that in post #123 (a short time ago today). On the issue of the Doppler effect, there would be no change at all in frequency of the light for anyone co-moving with the clock, so it's only a stationary observer that would see the colour of the light change. The light in the moving clock would be produced at a lower frequency as the mechanism producing it would be slowed by movement through space, but it would also vary, becoming more blue when the light pulse is on the longer forward paths through its clock and more red when moving rearward on the short paths. The speed of the light through space would always be c though - the frequency changes cannot affect that.

...But it seems to me that, without doppler effect, even if light would take more time between the mirrors, the distance between the tics would always be measured the same whatever the speed, which means that a clock that has traveled during a certain time would show exactly the same elapsed time than one at rest.

If you look at my interactive MMX diagrams, you can see the stationary apparatus on the left with red dots representing light pulses moving through it, and you can see that the ticks are faster on it that they are on the moving version of the apparatus shown to the right of it. The MMX is directly equivalent to a pair of light clocks set perpendicular to each other, so the time taken for the light to leave the semi-silvered mirror and to return to it from one of the mirrors at the ends of the arms can be taken as a clock tick. The same slowing of ticks would occur for moving the apparatus in the opposite direction, so if you were to imagine it stationary next to the one on the left that's already shown as being stationary, and then move it to the right for a while, then stop it, then move it back to where it started, and then stop it there, you would be able to count the ticks for the stationary one and for the one that moves, and you'd then see that the one that moves counts up fewer ticks than the stationary one as a result of its movement. You should also be able to see that the light never runs slow in either clock, so there is no slowing of any real time - there is only a slowing of clocks due to the greater distances that light has to travel on a moving clock per tick. All functionality of moving things is slowed for the same reason.

[David Cooper]

tA=0.75s

tB=1.5s

thinks that's correct now?

What are those numbers supposed to represent?

Anyway David, what i have been trying to explain to you is that it means nothing.  There is no contraction of space or the carriage. There is only objectively  a variance in distance that gives you a variance in tick rate.

There is a reduction of ticks for moving clocks compared to stationary clocks, and the slowing of clocks in the real universe matches up with the increased distances that light travels on light clocks set perpendicular to their direction of travel and with length-contracted light clocks aligned with their direction of travel, but until you can produce the right numbers or read what's happening on interactive diagrams correctly, you'll continue to misunderstand all of this and I don't have the time to drag you kicking and screaming through the rest of it if your learning speed is going to stay so low. You need help from AGI and I'd rather put the time into building that AGI.

[guest39538]

Quite clearly you think a clock is more than it is.   I know what a light clock is but quite clearly you ignore the observer affect by adding mirrors and such, setting the parameters to fit the ''story'' without considering the what at best you have in terms of objective reality.  The clock does not tick slower to begin with, we can work out the extra distance the light needs to travel and adjust accordingly to maintain the same rate of tick. i.e 1 second would not be equal to 1 second unless we calculated the difference to synchronise the difference.
I.e one clock would be measuring 1 second while 1 clock was measuring 1.2 seconds. The duration and length of a second then remaining the same with no contraction needed .

Picture this. We have two light clocks sitting next to each other, and let's align them the same way as each other. We move one of the light clocks away from the other and it ticks less often than the stationary one because of the increased distance light has to travel in the moving clock to complete each tick. We stop moving that clock and see it return to ticking at the same rate as the clock that never moved. Next, we move it back again to reunite it with the stationary clock, and while it's moving back it ticks more slowly again. Once we have brought it to a halt next to the clock that never moved, the two clocks tick in sync with each other again. If each clock has a counter, as clocks usually do, they might both have been set to zero at the start of the experiment. Now one of the clocks has a higher reading on it than the other because it has recorded more ticks. The clock that did all the moving has registered fewer ticks. The light in both clocks travelled exactly the same distance through space during the experiment, but it had to go further in the moving clock for many of the ticks and therefore couldn't complete as many ticks for its clock as the light in the stationary clock.

To claim that the clock that moved didn't tick more slowly than the one that stayed still the whole time is a nonsense - the different counts that they've racked up prove it.

The clocks do measure a different amount of ticks for the obviousness that you are changing  the distance.  You are doing two different distances so of course it measured different ticks, however it still means nothing, there is still no contraction and all you have shown is light takes more time to travel a further a distance which is quite obvious.   So what is your big reveal that is going to wow me? 

[guest39538]

tA=0.75s

tB=1.5s

thinks that's correct now?

What are those numbers supposed to represent?

They represent the time it takes light to travel ebb and flow in a moving carriage.

Anyway David, what i have been trying to explain to you is that it means nothing.  There is no contraction of space or the carriage. There is only objectively  a variance in distance that gives you a variance in tick rate.

There is a reduction of ticks for moving clocks compared to stationary clocks, and the slowing of clocks in the real universe matches up with the increased distances that light travels on light clocks set perpendicular to their direction of travel and with length-contracted light clocks aligned with their direction of travel, but until you can produce the right numbers or read what's happening on interactive diagrams correctly, you'll continue to misunderstand all of this and I don't have the time to drag you kicking and screaming through the rest of it if your learning speed is going to stay so low. You need help from AGI and I'd rather put the time into building that AGI.

 

Slowing clocks?  what on earth are you talking about?  the only slowing down is the subjective parlour trick you are trying to introduce which means nothing and shows nothing.

p.s think my numbers should be 0.67s and 2.01s  sorry keep changing my mind

[guest39538]

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Yes I made this way up.

[guest39538]

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[Le Repteux]

The light in the moving clock would be produced at a lower frequency as the mechanism producing it would be slowed by movement through space, but it would also vary, becoming more blue when the light pulse is on the longer forward paths through its clock and more red when moving rearward on the short paths.

Could you add the doppler effect to your simulation with the contracted distance? Then we could see with our own eyes if the system tics more slowly than the source or not. Your source could emit a dot each second for instance.

[David Cooper]

Could you add the doppler effect to your simulation with the contracted distance? Then we could see with our own eyes if the system tics more slowly than the source or not. Your source could emit a dot each second for instance.

It would be better to make a new diagram for that rather than messing up the existing one, but it would also take a lot of time and effort to create it, and I'd rather leave that work to someone else. JavaScript is not good at doing graphics and it takes a lot of trial and error to build things and line everything up properly, with each browser needing a different alignment set up for it and with browser updates sometimes messing the alignment up so that the dots are no longer where they're supposed to be. The light pulses are done using a full stop ("."  - yes, it's a piece of punctuation) and the MMX apparatus is an image file because the only other way of drawing lines is to use lots of "_" and "|" characters and then write code to move each one around if the object has to move. There was an old system for doing graphics along with JavaScript, but it was discontinued and the older versions of my programs became obsolete overnight. The replacement system was SVG which is a horrific bloated mess that I decided not to learn, not least because it too could have been discontinued at the drop of a hat too, but also because when I tried out the example code on the site behind SVG it didn't work on my machine at all, so I came up with ways of using JavaScript itself to do graphics in combination with HTML and then rewrote my programs to work that way. Even so, it's so much effort that I really don't want to have to write another graphics program in JavaScript ever again (or indeed any program in JavaScript as the language does not agree with the way I'm comfortable writing code - it too is a nightmarish mess).

But is it really necessary? You can already see ticks in the existing interactive diagrams - a tick occurs when the light diverges from the semi-silvered mirror and another occurs when the light gets back to the semi-silvered mirror. If the diagram illustrated a pair of light clocks, the emitter and receiver would both be located at the semi-silvered mirror and a new pair of light pulses would be sent out along the arms as soon as the first one returned. You should be able to imagine that happening without having to see it. You can already see how long the delay is between ticks and compare it against the action on the stationary MMX apparatus on the left of the screen.

I can see though that it would be nice to have a program that puts out lots of light pulses per cycle rather than just one because you'd be able to get a feel for how often they pass a stationary point (or a point on a moving object), but I can already imagine that such a program would immediately lead to certain people complaining about the rate at which the light pulses are generated because they'll insist that they should be sent out one for every second of absolute time instead of the slowed time of the moving device, so then another, smaller co-moving light clock would have to be programmed in to trigger the release of each light pulse on the larger clock, and then a string of other misplaced objections would follow while I spend weeks or months modifying the program to try to illustrate more and more things that they will simply continue to misunderstand or fail to see regardless. I'm not prepared to do that - it's a job for AGI to do and not one for me.

[guest39538]

Is this correct now? my final answer

ruler1.jpg (110.31 kB . 1765x505 - viewed 4694 times)

All the even numbers would be equal in length, and all the odd numbers would be equal in length.


2  0.66666666666
3  1.33333333333
4  0.66666666666
5  1.33333333333
6  0.66666666666
7  1.33333333333
8  0.66666666666

2+3=2.s

20cm + 40 cm = 60cm

At rest it takes 2 seconds for a round trip of 60cm.

In motion it takes 2 seconds for a round trip of 60cm.

Wheres your problem?

[David Cooper]

The clocks do measure a different amount of ticks for the obviousness that you are changing  the distance.  You are doing two different distances so of course it measured different ticks, however it still means nothing, there is still no contraction and all you have shown is light takes more time to travel a further a distance which is quite obvious.   So what is your big reveal that is going to wow me?

Five points:-

(A) A moving clock records fewer ticks than a stationary one in a given length of time, so it is not recording time, but is merely counting cycles. That is what all clocks do.

(B) A light clock aligned perpendicular to its direction of movement therefore records fewer ticks than a stationary clock.

(C) An uncontracted light clock aligned with its direction of travel (rather than perpendicular to it) will record fewer ticks than a light clock co-moving with it which is aligned perpendicular to their direction of travel, but you can't see that yet because you've only attempted the maths for one of the two cases, and you haven't even got that right.

(D) A correctly length-contracted light clock aligned with its direction of travel will record the same number of ticks as a light clock co-moving with it which is aligned perpendicular to their direction of travel.

(E) The null result of the MMX shows that the real universe length-contracts things in their direction of travel.

However, you won't agree with half of that because you still can't apply valid methods with the maths, even though I've shown you how to do it all for the part you keep tripping up on. I've tried to get you to the point where you can fit correct numbers to the non-perpendicular case, but you still aren't there and I think you're doing everything you can to avoid getting there. We haven't even started on the perpendicular case, so you still a long way from the point where you can compare the results and see that they don't tick at the same rate as each other. It doesn't look as if you're capable of completing the journey through this stuff, and the reason is most likely down to you not wanting to know that your position on it has been wrong all this time, based on a long string of errors in your thinking. I don't have time to fix it all for you - it's like pushing a car while the person sitting in it is applying the brakes as hard as they can.

[guest39538]

The clocks do measure a different amount of ticks for the obviousness that you are changing  the distance.  You are doing two different distances so of course it measured different ticks, however it still means nothing, there is still no contraction and all you have shown is light takes more time to travel a further a distance which is quite obvious.   So what is your big reveal that is going to wow me?

Five points:-

(A) A moving clock records fewer ticks than a stationary one in a given length of time, so it is not recording time, but is merely counting cycles. That is what all clocks do.

(B) A light clock aligned perpendicular to its direction of movement therefore records fewer ticks than a stationary clock.

(C) An uncontracted light clock aligned with its direction of travel (rather than perpendicular to it) will record fewer ticks than a light clock co-moving with it which is aligned perpendicular to their direction of travel, but you can't see that yet because you've only attempted the maths for one of the two cases, and you haven't even got that right.

(D) A correctly length-contracted light clock aligned with its direction of travel will record the same number of ticks as a light clock co-moving with it which is aligned perpendicular to their direction of travel.

(E) The null result of the MMX shows that the real universe length-contracts things in their direction of travel.

However, you won't agree with half of that because you still can't apply valid methods with the maths, even though I've shown you how to do it all for the part you keep tripping up on. I've tried to get you to the point where you can fit correct numbers to the non-perpendicular case, but you still aren't there and I think you're doing everything you can to avoid getting there. We haven't even started on the perpendicular case, so you still a long way from the point where you can compare the results and see that they don't tick at the same rate as each other. It doesn't look as if you're capable of completing the journey through this stuff, and the reason is most likely down to you not wanting to know that your position on it has been wrong all this time, based on a long string of errors in your thinking. I don't have time to fix it all for you - it's like pushing a car while the person sitting in it is applying the brakes as hard as they can.

I do not think you viewed my latest last post.  My numbers are correct.

[guest39538]

FIFY :  Only measure the time using the front detector and count in 2's.

P.s I did warn David to be armed with more than subjective ''parlour tricks''.   I objectively ''see'' no problems in my results conclusion.

added ; my scale I am using is

30cm:299 792 458 m

speed ratio 2:1

[GoC (OP)]

You still haven't taken in the scale of your monumental error! The 7.4641t is a time and not a distance. The distance is that time multiplied by 10 (which comes from the 10cm distance). The train moves over seventy four cm before the light reaches the front mirror.

There is no such thing as time. There is only one distance measured by another distance that cycles. Light will hit the front mirror in 7.461 cm in 10 cm length. It will hit the front mirror in 0.74671 n a 1 cm length. It is a ratio. A 100 cm cell length will be 74.61 ratio. There is no fixed time only a fixed ratio to c for energy of motion. In a frame time measurement is the same ratio for a cm as a km.

I am confused to my monumental error?

[Le Repteux]

You can already see how long the delay is between ticks and compare it against the action on the stationary MMX apparatus on the left of the screen.

Of course, if we wait till the dots hit the detector to send a new one, then there will be more time between the dots on the right diagram, but if we send them at one second interval on both diagrams, it seems to me that they will be detected at the same frequency on both detectors, because there will be no doppler effect to alter the frequency. Time depends on the frequencies of vibrating or rotating bodies, not on the distance they travel through the fabric of space during their rotation or vibration, so why would it depend on the distance light travels instead of depending on its frequency? It is not the distance the earth's surface travels in space that determines the sidereal day length, it's when the same star passes at the zenith. It is not the distance traveled by a pendulum that determines the tics of a clock either, it's when it passes in the middle of its course.

[David Cooper]

I do not think you viewed my latest last post.  My numbers are correct.

If they're correct, you should be able to insert them into the right places [between square brackets] in the following list:-

(1) Length of vehicle = d [= 299,792,458m]

(2) Time for light to travel distance d = t [= 1s]

(3) Time for light to make round trip lengthways when vehicle at rest = 2t [= 2s]

(4) Time for light to make first part of trip when vehicle moving at 0.5c = 2t [= 2s]
(Front of vehicle was ahead of light by d and moving at 0.5c while light is moving at c, so light is gaining on front of vehicle at 0.5c and will take 2t to catch it.)

(5) Distance vehicle has moved by this point = d [= 299,792,458m]
(The light moved 2d and the vehicle moved half that.)

(6) Distance light has moved by this point = 2d [= 2 x 299,792,458m]

(7) Time for light to make second part of trip = 2/3t [= 2/3s]
(This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)

(8 ) Distance vehicle has moved during the time the light was coming back = 1/3d [= ...]

(9) Distance light has moved during second part of trip = 2/3d [= ...]

(10 ) We now have a round trip for the light completed in 2 2/3t [= ...s]. The light has moved 2 2/3d [= ...] through space. The vehicle has moved a total of 1 1/3d [= ...], which is half the distance the light travelled, and that's no surprise as the light was moving twice as fast as the vehicle.


You say that your latest numbers are correct, but I haven't seen you produce the right answers yet to put into (8 ) and (9), and those are the numbers you should be calculating. Also, you still haven't told me whether you agree with the number in square brackets for (7). Until these [= ...] parts remain unfilled with your numbers, you have not completed your assignment.