- #176
harrylin
- 3,875
- 93
roineust said:harrylin,
I don't understand exactly what you mean by cross trip, but if you mean from one bank of the river, to the other bank, and the current speed is 2kph, and the boat speed is 5kph, and the river width is, say 1km, then I think the time to cross to the other bank would be:
1/(root((2^2)+(5^2)))=1/(root(29))= ~0.185 hours= ~11 minutes.
Would that be correct?
Roi.
I don't think that I wrote "cross trip"
But you already answered another question that is relevant for MMX.
And yes your answer is almost right - it's just trigonometry.
Funny enough, it's a question that Michelson also had wrong the first time that he calculated it:
The boat has a constant speed of 5 km/h, let's call that velocity c.
Now replace the boat by light.
Also, replace the river bank by the MMX apparatus (moving at velocity v).
Then, in 1881 Michelson calculated 0.20 hours for your example! But when he did his famous experiment in 1887, he calculated it correctly and he found more than 0.20 hours. Do you understand why it will take longer when there is a strong river current?
But I wrote "How do you calculate the time that the motorboat needs to go from one place to another? and back?" and "Yes that's right - for a boat sailing along with the river. And how long does the return trip take?".
The end question is: how long will it take to go from A to B downstream and immediately back upstream to A?
If you have that, and you solved the little error in your calculation here above, then you can do Michelson's calculations. And after that, you may have the basic knowledge to follow the things we discussed with you.
Last edited: