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# Foo Forum

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science question
goofy grandpa
dibbern2

Msg#: 3921231 posted 10:29 pm on May 27, 2009 (gmt 0)

My grandson has posed this question to me and I'm too science-dumb to give him an answer. But I did say "I know just where to ask!" So here goes:

Pretend there is a train that has very tall cars, say, 30-40 feet. Also pretend (assume) the train can go very fast. 500 MPH sounds good.

You are in one of the cars. You paint a bulls-eye on the floor of the car. You set-up a ladder that allows you to climb to the top of the car. From there, you position yourself to drop a golf ball exactly over the bulls-eye on the floor below.

Will the ball land exactly below where you drop it, or will it land a little behind the bulls eye, loosing delta while it falls and car is rushing forward at 500 mph?

I think I know that the ball will hit the bulls-eye with no effect from the car moving forward, but common sense tells me it will land behind.

Of course, if it did land behind the mark, then things dropped from any great height would always land a little west of straight down as the earth turned during the fall, wouldn't they?

Leosghost

Msg#: 3921231 posted 10:36 pm on May 27, 2009 (gmt 0)

it will land a little behind..
this is very close to some of the thinking that led Albert to the theory of relativity ..
His example was a beam of light shone across a falling elevator car.

edit typo

[edited by: Leosghost at 10:37 pm (utc) on May 27, 2009]

LifeinAsia

Msg#: 3921231 posted 10:46 pm on May 27, 2009 (gmt 0)

Assuming the car is airtight, if the velocity of the train is constant, it should hit the bulls eye. If you drop it while the train is accelerating or decelerating, it should miss the mark (how much and how far depends on the rate of acceleration/deceleration and air pressure inside the car). (This is similar to the example under Newton's first law found here [en.wikipedia.org].)

 then things dropped from any great height would always land a little west of straight down as the earth turned during the fall, wouldn't they?

No, because the object would be have "turned" the same amount.

[edited by: LifeinAsia at 10:59 pm (utc) on May 27, 2009]

kaled

Msg#: 3921231 posted 10:57 pm on May 27, 2009 (gmt 0)

The Earth is a giant train moving at many thousands of miles an hour around the sun and and rotating at about 1000 mph about its axis (at the equator). Yet, for most practical purposes, it may be considered to be stationary.

If the train is moving in a straight line, with uniform speed, on horizontal track, and is airtight, the object will fall vertically relative to an observer on the train.

Kaled.

Msg#: 3921231 posted 8:44 am on May 28, 2009 (gmt 0)

 If the train is moving in a straight line, with uniform speed, on horizontal track, and is airtight, the object will fall vertically relative to an observer on the train.

Rephrased with a tighter condition on the fist three qualifiers - If the train moves at a uniform velocity [en.wikipedia.org], and the container is airtight, the object ill fall vertically relative to an observer on the train (hitting the bullseye)

Leosghost, relativistically you are right. However, at the speeds concerned, air turbulance will have more influence than relativity calculations. At low energies, Newton is more than adequate.

[edited by: lawman at 10:14 am (utc) on May 28, 2009]
[edit reason] Fix Style Code [/edit]

Leosghost

Msg#: 3921231 posted 10:53 pm on May 28, 2009 (gmt 0)

Everything is relative ~:)

kaled

Msg#: 3921231 posted 11:26 pm on May 28, 2009 (gmt 0)

Uniform speed in a straight line == Uniform velocity.

I used a description that someone would understand without knowing the scientific definition of velocity.

Incidentally, whilst relativistic physics is not my forte, the speed of light (in a vacuum) is a universal constant and does not ever change with respect to the observer. i.e. however fast the train is moving, an experiment to measure the speed of light on the train will always yield the same result in every direction, up down, forwards, backwards, etc. (and light will still travel in straight lines).

Kaled.

Leosghost

Msg#: 3921231 posted 11:44 pm on May 28, 2009 (gmt 0)

Gravity ..and Heisenburg and some peoples cats might wish to bring the weight of their arguments and observations to bear or at least exert some influence upon the discussion .

Leosghost

Msg#: 3921231 posted 11:52 pm on May 28, 2009 (gmt 0)

the path of light is easy to bend ..sufficient mass or liquid refraction ( air can be considered to be a liquid ) or even spectacles ( or mucus upon their cornea ) on the nose of the beholder can bend ( or repath light ) light ..the velocity of light is indeed a constant ..however ..where it goes ,what it does and what it illuminates or does not and how that is interpreted are all ..as I said ...relative..

Leosghost

Msg#: 3921231 posted 11:57 pm on May 28, 2009 (gmt 0)

nothing moves in a straight line ..everything is falling ..in a curve ..but from where to where is the interesting part ..to paraphrase Laurie Anderson ..

Leosghost

Msg#: 3921231 posted 12:04 am on May 29, 2009 (gmt 0)

 with respect to the observer.
the observer can never be static in relation to the observed phenomena..their masses change and the distance between a dropped object and the "dropper" change ..etc etc ..even simple gravity behaves ( more or less from where we are ..most of the time ) in an inverse square ratio dependant upon the distance between the relative masses ..
g1smd

Msg#: 3921231 posted 12:12 am on May 29, 2009 (gmt 0)

*** everything is falling ..in a curve ***

...rotation around:
- Earth's own axis
- the Earth-Moon barycentre
- the Sun
- the galaxy
- the local group
- the universe.

[edited by: g1smd at 12:13 am (utc) on May 29, 2009]

stajer

Msg#: 3921231 posted 12:12 am on May 29, 2009 (gmt 0)

 Rephrased with a tighter condition on the fist three qualifiers - If the train moves at a uniform velocity, and the container is airtight, the object ill fall vertically relative to an observer on the train (hitting the bullseye)

Just because I am a stickler - the train car must be a perfect vacuum, not just air tight.

The answer here is the same reason jumping upward a moment before a free falling elevator hits the ground will not save you.

lawman

Msg#: 3921231 posted 2:02 am on May 29, 2009 (gmt 0)

 The answer here is the same reason jumping upward a moment before a free falling elevator hits the ground will not save you.

But I remember an old Bugs Bunny cartoon where he was in an airplane careening toward earth but was saved when the plane ran out of gas just a few feet before it hit the ground. How do you explain that?

Woz

Msg#: 3921231 posted 2:09 am on May 29, 2009 (gmt 0)

Air Brakes of course! (I saw that one too Lawman.)

Onya
Woz

[speeliong]

[edited by: Woz at 2:50 am (utc) on May 29, 2009]

willybfriendly

Msg#: 3921231 posted 4:35 am on May 29, 2009 (gmt 0)

 Of course, if it did land behind the mark, then things dropped from any great height would always land a little west of straight down as the earth turned during the fall, wouldn't they?

And, they do! Long range artillery fire (as from battleships) must take this Coriolis effect into account.

The Brits failed to adjust for changes moving from northern to southern hemisphere in the Faulkland Island war resulting in some very inaccurate gunnery at the beginning.

Another example would be the Paris gun used by the Germans in WWI (from a range of ~75 miles).

It is the rotation of the earth that causes the phenomon. The projectile is traveling in a straight line while the earth turns beneath it.

A train would be subject to the same effects, since a "straight" line really is following the curve of the earth. The ball's trajectory could be computed via Newtonian physics using a vector matrix. It is a combination of the momentum imparted by the earth's spin (think cetrifugal force) throwing the ball in a straight line away from the earth, the forward momentum of the ball (equal to the train's speed and direction at the moment it is released) and gravity pulling the ball straight towards the center of the earth (at the moment it is released).

The ball would not hit the target, although it is simplistic to say that it would land a bit behind it.

kaled

Msg#: 3921231 posted 7:51 am on May 29, 2009 (gmt 0)

Newton's first law...

A object will continue in its state of rest or uniform motion unless acted upon by some external force.

There is no horizontal force, therefore there is no horizontal acceleration, therefore the object will travel with precisely the same horizontal velocity as the train.

 the observer can never be static in relation to the observed phenomena

In all such experiments, the observer is defined as stationary (because direct measurements can only be made relative to oneself - the observer). Gravity does obey the inverse square law (provided the objects are spheres rather than long cylinders) however, that is wholly irrelevant. For the purpose of the discussion, gravity is uniform (because the train is moving horizontally). We could also introduce magnetism, and electrostatic charge, but that would be equally nonsensical.

Kaled.

Msg#: 3921231 posted 8:45 am on May 29, 2009 (gmt 0)

This is fun!
 Just because I am a stickler - the train car must be a perfect vacuum, not just air tight.

Not true! If the cabin is airtight, then the air will be traveling at the same rate as the train- thus not moving at all relative to the experiment. Conceptually, dropping a ball at a target in a stationary environment (relative to the surface of the Earth), will result in a hit. Same with an airtight cabin.

 I used a description that someone would understand

Pah! The unwashed masses don't deserve to understand <--- This is clearly a joke, not a breach of TOS
 Coriolis effect

Not relavant, as the Rate Of Change of latitude of the target is equal to the ROC of latitude of the object.

With a nod to Leosghost post #3922083 on Heisenburg and cats...

You relativists might as well argue that since you know the precise velocity of the target, you can't possibly know where it will be when the ball is due to hit it. Its simply not the right scale for that argument is to apply.

Of course, if the train was properly sealed, the ball would have both and neither hit and/or missed the target.

willybfriendly

Msg#: 3921231 posted 1:46 pm on May 29, 2009 (gmt 0)

 There is no horizontal force, therefore there is no horizontal acceleration, therefore the object will travel with precisely the same horizontal velocity as the train.

That would be true if the earth were flat!

The train is following a curved path (the surface of the earth). The object will proceed in a straight path as the train curves away from it.

 ...the Rate Of Change of latitude of the target is equal to the ROC of latitude of the object.

Again, only if the earth were flat. The "horizontal" vector of the train is in fact following a curved path. Once the object is released its vector will be a straight path.

We are, after all, talking about a three dimensional space.

kaled

Msg#: 3921231 posted 4:11 pm on May 29, 2009 (gmt 0)

 Once the object is released its vector will be a straight path

If you really want to be that silly and pedantic, the object would travel in a sub-orbital curve after it is released. A curved trajectory would be apparent even to an observer on the train, provided the instrumentation was sufficiently accurate. However, since there is no information available as to speed, latitude, and direction, it is reasonable to ignore all effects that would depend on these variables.

Kaled.

Msg#: 3921231 posted 4:54 pm on May 29, 2009 (gmt 0)

More than reasonable. Its all about scale.

Whats the angular velocity of the fastest train? Negligible. Thus, any impact of any factor relying on this will can be ignored.

LifeinAsia

Msg#: 3921231 posted 5:04 pm on May 29, 2009 (gmt 0)

 But I remember an old Bugs Bunny cartoon where he was in an airplane careening toward earth but was saved when the plane ran out of gas just a few feet before it hit the ground. How do you explain that?

Obvious- no gas means no more acceleration. No acceleration = stand still.

Don't you know that the normal laws of physics don't apply to cartoon characters? Honestly, what were you thinking?

akmac

Msg#: 3921231 posted 7:47 pm on May 29, 2009 (gmt 0)

 There is no horizontal force, therefore there is no horizontal acceleration, therefore the object will travel with precisely the same horizontal velocity as the train.

Wouldn't the locomotion of the train be considered an external force?

As soon as the object is released, the golf ball will begin horizontal deceleration, losing momentum as it is no longer being directly acted upon by the power of the train. (apart from the air-which is also being acted upon by the locomotion of the train)

Ball hits behind the bulls eye.

I'll also argue that the ball will bounce and roll towards the rear of the train, eventually coming to rest at the back wall.

Unless of course our happy experiment is abruptly halted due to the failure of our myopic engineers to account for that tunnel...

Msg#: 3921231 posted 8:28 pm on May 29, 2009 (gmt 0)

 Wouldn't the locomotion of the train be considered an external [b]force[b]

Nope. Velocity (locomotion)* is not a force. Force acts on an object to ACCELERATE it. No (horizontal) acceleration here, so no force present.

As the cabin is defined as airtight, there is no horizontal drag on the ball, thus no horizonal force, thus no horizonal movement. It's straight down, relative to the cabin.

* The train's engine provides sufficient force to overcome the friction of the train through the air. Net force of engine plus drag is zero.

LifeinAsia

Msg#: 3921231 posted 8:35 pm on May 29, 2009 (gmt 0)

 Wouldn't the locomotion of the train be considered an external force?

Nope- no acceleration, so no external force. Except for gravity, which is pulling it downwards.

 the golf ball will begin horizontal deceleration

Nope- there is no force decelerating it.

 our happy experiment is abruptly halted due to the failure of our myopic engineers to account for that tunnel.

Or the failure of the engineers to account for the lack of a tunnel. SPAT! :)

willybfriendly

Msg#: 3921231 posted 9:19 pm on May 29, 2009 (gmt 0)

 It's straight down, relative to the cabin.

It is straight down relative to the center of the earth's mass.

The ball will miss the target...

We have been asked to assume an initial velocity of 500 MPH (roughly 224 meters per second) and a drop of 40 feet (a bit over a second of free fall). From the moment of decoupling the system the train will traverse a bit over 224 meters of the arc of the earth's diameter while the ball will travel a bit over 224 meters along the tangent of the arc. Been to long since I have done trig, but I would guess a simple solution would prove a miss of several centimeters.

arieng

Msg#: 3921231 posted 9:31 pm on May 29, 2009 (gmt 0)

willybfriendly, i think you just nailed it.

LifeinAsia

Msg#: 3921231 posted 9:43 pm on May 29, 2009 (gmt 0)

The difference would be so negligible that I doubt you could measure it. Certainly not centimeters.

grandpa

Msg#: 3921231 posted 10:04 pm on May 29, 2009 (gmt 0)

to dibbern2's grandson, learning to build decent websites is a whole lot easier than debating physics with a bunch of webmasters :) I say the ball misses the mark, and it really doesn't matter why to me - it's just one of those things that I either intuitively know or willingly accept.

dibbern2

Msg#: 3921231 posted 11:15 pm on May 29, 2009 (gmt 0)

Okay, I admit it: now I'm really confused.

But I have enjoyed the conversation.

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