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Adding Speeds

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Dr C Dodgson
1015717.  Fri Aug 09, 2013 2:44 pm Reply with quote

If you're traveling along at 2 mph while holding a ball, then you chuck the ball forward at 2 mph, how fast is the ball traveling (ignore air resistance)?

Of course the obvious answer is 4mph. But it's wrong. We think of adding speed by doing Vector1 + Vector2 = Result. That's the Newtonian way of doing things. And it's a good approximation. Good enough to get us to the Moon in fact. But it's still wrong for all that.

The correct formula uses relativity and is vector1 + vector2 divided by (1 + (V1 + v2/C). So instead of 4mph, the ball is actually going 3.9999999761407696025006286758089551099972853564134485 mph.

 
djgordy
1015722.  Fri Aug 09, 2013 2:54 pm Reply with quote

Have you ever tried chucking a ball to the Moon?

 
suze
1015735.  Fri Aug 09, 2013 3:56 pm Reply with quote

How often have you been able to measure both your traveling speed and the speed at which you throw the ball with sufficient accuracy that it is meaningful to quote their combined speed to eight places of decimals?

 
Dr C Dodgson
1015738.  Fri Aug 09, 2013 4:46 pm Reply with quote

djgordy...you've never heard of a moonballer?

suze, Not frequently. However the equation holds true if general relativity holds true. And general relativity has been measured down to, as Brian Cox memorably put it (on QI or some other program), the arc distance between Pluto's north and south poles....as seen from earth. I'm not entirely sure how you'd put that in #s, but I think we can at least say that the margin of error is very, very, very small.

 
Posital
1015741.  Fri Aug 09, 2013 5:22 pm Reply with quote

Air resistance isn't the only other force you'd need to account for.

You'd need to account for gravity (if not the earth, then their body/hand). As well as any light (or any other method of position/speed measurement).

And if our Boojum friend had been doing his homework - there would also be a nod to the uncertainty of that speed/position measurement as noted by mr heisenberg.

And most tellingly, no account has been taken of newton's third law (which still holds even in these relativistic times). As the force to propel the ball (the chuckee) would result in an equal and opposite force slowing down the chucker.

Each of which would mean that the number of decimal places would be lost among the error bars or inaccuracies in the above calculation.

But it's good as a simple thought experiment.

Anyway, I like trains better.

Welcome Lutwidge.

 
PDR
1015742.  Fri Aug 09, 2013 5:22 pm Reply with quote

Well if you're going to be pedantic (you started it!) your suggested answer is wrong, in many ways.

Firstly you'd have to recognise that you're talking about "velocity" when adding vectors because "speed" is a scalar (but that's just nitpicking).

Secondly you'd need to define whether you're talking about the average speed of the ball (either total distance travelled divided by total time, distance travelled per unit time or time taken to travel a unit distance) or the instantaneous velocity of the ball (instantaneous speed being an invalid concept, and Heisenberg uncertainty constraints notwithstanding).

Thirdly since you're using a relativistic analysis you need to define whether it is velocity with respect to an outside observer, the thrower or the ball. Each will yield a different number.

Fourthly you need to establish whether this act occurs in free space or in a gravity well. If the latter then you'd need to account for the trajectory of the ball and its consequent profile through different gravitational field strengths - the time experienced by the ball will vary at different heights in the gravitational field.

Fifthly you'd need to define whether you're talking about the straight-line progress velocity of the ball or its velocity along its trajectory - these will also yield different numbers where there is any gravitational field.

Welcome to QI, BTW!

PDR

 
Sparkyweasel
1015934.  Sat Aug 10, 2013 6:04 pm Reply with quote

Dr C Dodgson wrote:
If you're traveling along at 2 mph while holding a ball, then you chuck the ball forward at 2 mph, how fast is the ball traveling?

The answer is on the question, if you chuck the ball at 2 mph, it travels at 2mph (to start with).

:)

 
Posital
1016127.  Sun Aug 11, 2013 11:28 pm Reply with quote

lol - gets my vote...

 
Paulmarkj
1203428.  Wed Aug 31, 2016 8:47 am Reply with quote

"The correct formula uses relativity and is vector1 + vector2 divided by (1 + (V1 + v2/C). So instead of 4mph, the ball is actually going 3.9999999761407696025006286758089551099972853564134485 mph."

The c should be c squared: (1 + (V1 + v2/(c*c))

So the answer would be 3.99999999999999996431472

 
PDR
1203544.  Thu Sep 01, 2016 7:47 am Reply with quote

Differing from 4 by a whopping 0.000000000000000892132% which, by a curious coincidence, is almost identical to the rate of Tax Apple wanted to pay last year.

Spooky...

PDR

 
dr.bob
1203889.  Mon Sep 05, 2016 8:03 am Reply with quote

Dr C Dodgson wrote:
If you're traveling along at 2 mph while holding a ball, then you chuck the ball forward at 2 mph, how fast is the ball traveling (ignore air resistance)?


If you're going to bring relativity into the mix, then you need to rephrase the question by specifying relative to which observer you're measuring the speed of the ball.

 
PDR
1203891.  Mon Sep 05, 2016 8:35 am Reply with quote

Ahem thirdly...

PDR

 

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