[geordief]

what would be the external observer's measurement of the speed of light within the craft?  My instinctive thought would be that she would measure both as c. which would mean that the light was perceived as stationary relative to the craft.

No, think about it a little longer and I’m sure you will get it.

Surely an external observer could only measure the speed of light within the craft in an indirect way.

In any direct measurement it would just be c.

If two object's are mutually receding at .9  from a third observer then that observer can say they are separating at up to 1.8c  but each object will only see the other moving at .9c (if that is relevant)

[Bill S (OP)]

If two object's are mutually receding at .9  from a third observer then that observer can say they are separating at up to 1.8c  but each object will only see the other moving at .9c (if that is relevant)

I'm unclear about the set-up, here.  Are the two objects receding from each other; both travelling at 0.9c? 
Where is the observer relative to the two objects?
Are the two objects receding from the observer?

[jeffreyH]

You cannot have the ship traveling at c since time dilation in that frame of reference would be infinite. No force can be applied in a frame of reference that has infinite time dilation. This reference frame can only be inertial. This is ignoring the fact that infinite energy would be required to accelerate the ship to the speed of light. This is according to theory, although since we have infinite time dilation at that speed the energy can never actually reach an infinite value.

[Janus]

If two object's are mutually receding at .9  from a third observer then that observer can say they are separating at up to 1.8c  but each object will only see the other moving at .9c (if that is relevant)

I'm unclear about the set-up, here.  Are the two objects receding from each other; both travelling at 0.9c? 
Where is the observer relative to the two objects?
Are the two objects receding from the observer?

Assume observer 1 is at point A.  Object 2 is receding from point A to the East at 0.9c relative to point A and object 2 is receding from point A to the West at 0.9c relative to point A.
Observer 1 and object A will measure their respective relative velocity as being 0.9c
Observer 1 and object B will measure their respective relative velocity as being 0.9c
Observer 1 will measure the respective velocity difference between A and B's velocities as being 1.8c
Objects A and B will measure their relative velocity to each other as being ~0.994475c ( addition of velocities theorem)

Acknowledging that no craft could travel at c, but bowing to Einstein's thought experiment:  what would be the external observer's measurement of the speed of light within the craft?  My instinctive thought would be that she would measure both as c. which would mean that the light was perceived as stationary relative to the craft.

That depends on the direction the light is traveling.
If the ship is moving at 0.999c, relative to our observer, then he will measure the difference between light traveling in the same direction as the ship as moving at 0.001c with respect to the ship, but light traveling in the other direction will be moving at 1.999c with respect to the ship. (The light traveling from the face of the ship passenger to mirror will take longer to reach the mirror than the light reflecting off the mirror will take to make the round trip.)

[geordief]

If two object's are mutually receding at .9  from a third observer then that observer can say they are separating at up to 1.8c  but each object will only see the other moving at .9c (if that is relevant)

I'm unclear about the set-up, here.  Are the two objects receding from each other; both travelling at 0.9c? 
Where is the observer relative to the two objects?
Are the two objects receding from the observer?

Janus has described my set-up exactly(and clearly)

He also corrected my statement that the two objects would measure their relative separation at .9c to ~0.994475c.... (quite a large difference)

[yor_on]

"Recast Einstein’s thought experiment in terms of Galileo’s below-deck scenario: a sailor would know if the ship were moving at the speed of light because his reflection would vanish."

Why?

Einstein saw it the other way. So do I, so do all experiments.

[yor_on]

Janus, are you talking about a experimental setup 'at rest' with each other/itself, mounted inside a 'ship', it having a 'uniform motion', measuring the light of speed? If so, feel free to use https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment, and consider Earth the 'ship'. Ahh ok, three objects, then it makes sense. That's pretty simple actually, it's red shifts relative blue shifts as measured between them plus that the 'experimental setup' consisting of those ships no longer can be said to be 'at rest' with each other/itself
=

ahh two, not three. Doesn't matter really:)
I looked at the figures first, that's what made me react.

[Bill S (OP)]

"Recast Einstein’s thought experiment in terms of Galileo’s below-deck scenario: a sailor would know if the ship were moving at the speed of light because his reflection would vanish."

Why?

Einstein saw it the other way. So do I, so do all experiments.

I expressed that badly.  My intention was to say that if, as Einstein reached light speed in his thought experiment, his reflection vanished from the mirror; then the same would apply to the sailor, so he would know he was moving.  Thus, the reflection would not vanish in either case.

[Bill S (OP)]

You cannot have the ship traveling at c ....

Nor any other massive object, apart from Einstein, and even he could do it only mentally.

[jeffreyH]

A useful subject to read up on is the velocity addition formula. While the mathematics may not look friendly it does include the Lorentz factor which should be familiar as gamma in relativistic equations.
https://en.wikipedia.org/wiki/Velocity-addition_formula

[Bill S (OP)]

Thanks Jeffrey. I spent some time and effort on that a few years ago. I think it's one of the few bits of maths I managed to grasp. I don't pretend to have it in my head, but I wrote myself a fairly simple explanation.