The arc of a released object

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The arc of a released object
« on: October 17, 2006, 12:12:08 AM »
Galileo observed that when an object was released and allowed to fall straight down, it would arc.  Believing it to be possibly wind, or a defect of his release, he repeated the experiement from different heights on still days, with more and more massively weighted balls (balls in each case) and observed that the higher he travelled, the more pronounced the arc.

The arc was always precisely the opposite of the sun's procession across the sky.  That is to say, the objects appeared to fall east, always, by some measure.

Why that was he conjectured, was that, in a rotating system, you have inertia and a constraint preventing the exterior bits from flying away.  This force, inertia, he hadn't had a specific name for.  Newton called it ctrifugal force, but it's just inertia.

THe further from the axis you travel, the faster your speed is, so long as you maintain the same RPMs.

An object has a larger cirucmference to travel along while maintaining the same rate of revolution, so naturally, goes faster.

This being the case, a cannonball released from the top of a high tower, is travelling faster toward the direction of the earths' rotation, assuming it's round and rotates, than at the middle of the tower.

If only by a small degree, but a degree large enough to be clearly seen when th eballs reached rest on the ground.

As the balls plummet, they are falling relative to things moving at the same rate of revolution, but being closer to the axis, are moving at a slower rate of speed.  The balls then appear to accelerate slightly before slownig their speed differential, and falling more or less straigt down.

You get a nice arc.

Now.  This is easily tested, wherever you ahve a tower you can climb and a cannonball you can find :)  BUt try it.

It is a readily observable phenomenon.

My question is, how is this explained in a flat earth scenario?  In a flat earth scenario, the ball being a weight on a disc, being released, should move toward the outside of the disc, that is to say, southward.

take a pice of cardboard, cut a circle out of it, put a pencil through it, then put a penny in the middle, and watch the penny move off.

Now, i fyou believe that the disc of the earth sits still, doesn't rotate, there should be no arc to massive falling spheres that can't be explained by wind.


At least, not using conventional physics.  So how is this accounted for in flat earth physics?

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dysfunction

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The arc of a released object
« Reply #1 on: October 17, 2006, 05:38:02 AM »
Welcome to the forum, and thank you for posting an original question. I'm not sure if that question has ever been asked here, and I have no answer for you, but I'm not a Flat-Earther, so you will have to wait until one comes by. In the meantime, enjoy the free bump.
the cake is a lie

The arc of a released object
« Reply #2 on: October 17, 2006, 05:46:11 AM »
centrifugal force doesn't exist  :D

by finding this one error in your argument, i have thus proved your entire theory inncorrect   :?  :lol:

course your 100% correct wen u refer to it as inertia... *sigh* forget my argument then  :o  and get someone with a brain to answer this question

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dysfunction

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The arc of a released object
« Reply #3 on: October 17, 2006, 07:09:48 AM »
Quote from: "woopedazz"
centrifugal force doesn't exist  :D

by finding this one error in your argument, i have thus proved your entire theory inncorrect   :?  :lol:


Where did he use the term "centrifugal force"?
the cake is a lie

The arc of a released object
« Reply #4 on: October 17, 2006, 07:18:01 AM »
Quote
This force, inertia, he hadn't had a specific name for. Newton called it ctrifugal force, but it's just inertia.

There you go!

As for the experiment, how exactly would it be observable? Would it be measured as a displacement from "straight down"? And if so, how are you measuring "straight down"?

The logistics of this type of experiment are bothersome.

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dysfunction

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The arc of a released object
« Reply #5 on: October 17, 2006, 07:26:52 AM »
Not too bothersome- find a level surface (using a level, of course), put up a tower of reasonable height, put a ruler down on the ground along the direction of the Earth's rotation, and drop a dense object with low air resistance (say a bowling ball) on a day with as little wind as possible. Set up a camera to record exactly where on the ruler the ball lands. If the Earth is rotating, the ball should land farther along the ruler than where it was dropped.
the cake is a lie

The arc of a released object
« Reply #6 on: October 17, 2006, 07:56:58 AM »
Quote from: "dysfunction"
Not too bothersome- find a level surface (using a level, of course), put up a tower of reasonable height, put a ruler down on the ground along the direction of the Earth's rotation, and drop a dense object with low air resistance (say a bowling ball) on a day with as little wind as possible. Set up a camera to record exactly where on the ruler the ball lands. If the Earth is rotating, the ball should land farther along the ruler than where it was dropped.

It seems like the level might be subject to the same inertial conditions as the bowling ball, and so level wouldn't really be level.

The arc of a released object
« Reply #7 on: October 17, 2006, 10:29:05 AM »
Quote from: "Unimportant"
Quote
This force, inertia, he hadn't had a specific name for. Newton called it ctrifugal force, but it's just inertia.

There you go!

As for the experiment, how exactly would it be observable? Would it be measured as a displacement from "straight down"? And if so, how are you measuring "straight down"?

The logistics of this type of experiment are bothersome.


They're pretty simple.  Aside from a plumb line, a level and square would also indicate "straight up" and "Straight down"

Simple carpenter's tools from ancient times are all that are required.  And the ability to measure how far east or west your point of release is at different heights, compared with one another, dould also suffice, without knowing up or down.  You would then just be able to measure relative distance the ball travelled eastward.

The arc of a released object
« Reply #8 on: October 17, 2006, 10:32:47 AM »
Quote from: "Unimportant"
Quote from: "dysfunction"
Not too bothersome- find a level surface (using a level, of course), put up a tower of reasonable height, put a ruler down on the ground along the direction of the Earth's rotation, and drop a dense object with low air resistance (say a bowling ball) on a day with as little wind as possible. Set up a camera to record exactly where on the ruler the ball lands. If the Earth is rotating, the ball should land farther along the ruler than where it was dropped.

It seems like the level might be subject to the same inertial conditions as the bowling ball, and so level wouldn't really be level.


why would it not be level?

If you keep the level at the same height, or if you drop it, it will still indicate up by the location of the air bubble.

In the case of a plum level, dropping a plumb level from a tower doesn't help anything... but subject to the same inertial forces, so is the frame it's suspended from, and the ground on which it sits, the observer.

All mas is subject to inertia.  How does that make a level inaccurate?

Why would it not indicate up and down?  Explain?

These are also questions not relevant to the original one.

How can flat earth science explain this repeatable and observable phenomenon, the easterly direction of freefalling spheres from great heights?

The arc of a released object
« Reply #9 on: October 17, 2006, 10:34:59 AM »
Quote from: "dysfunction"
Not too bothersome- find a level surface (using a level, of course), put up a tower of reasonable height, put a ruler down on the ground along the direction of the Earth's rotation, and drop a dense object with low air resistance (say a bowling ball) on a day with as little wind as possible. Set up a camera to record exactly where on the ruler the ball lands. If the Earth is rotating, the ball should land farther along the ruler than where it was dropped.


yeah, that's it in a nutshell.

ANd the distance along the ruler should be greater the higher the ball was released from.  And it shoul dbe repeated with a dropping machine, and multiple droppers, to factor out dropping technique, and repeated several times in each case, then statistically plotted.

It has been, of course, by galileo, and then by others since.  THe question is, how does a flat earth account for this?

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Erasmus

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The arc of a released object
« Reply #10 on: October 17, 2006, 10:45:35 AM »
Why did the chicken cross the Möbius strip?

The arc of a released object
« Reply #11 on: October 17, 2006, 10:52:35 AM »
Quote from: "Erasmus"


Constructing newton's laws in a rotating system and you see the same inertia.

The identical inertia.

It's lke calling inertia something else when its observed in a specific case.  That's stupid.

LIke saying "Those are clouds, unless they are over denmark, then we call them puffblobs.  ANd puffblobs, because they ave a different name are NOT LIKE CLOUDS"

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Erasmus

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The arc of a released object
« Reply #12 on: October 17, 2006, 11:11:09 AM »
Quote from: "Slorrin"
Constructing newton's laws in a rotating system and you see the same inertia.


Um, no, really, it's a force term.  Inertia and force are not the same thing.  Newton's 2nd Law in a rotating reference frame is

  F = ma  -  2m ω× v  -  m ω × (ω × r).

The last term, -m ω × (ω × r), is a vector pointing in the same direction as r (the vector from the center of rotation to the object in question) and has units of force (kg·m/s²) whereas inertia is a scalar quantity with units of mass or mass·distance.  It's definitely, unequivocally, a force.
Why did the chicken cross the Möbius strip?

The arc of a released object
« Reply #13 on: October 17, 2006, 01:00:11 PM »
Quote from: "Slorrin"
These are also questions not relevant to the original one.

They are relevent to whether this experiment is valid.

Would the plumbob not be deflected in the same direction as you claim the cannonball is displaced? If I accelerate along the surface of the earth with a plumbob in my hand, it trails behind me, even if I hold it at a constant height. If I understand your cannonball claim, you are saying it displaces eastward for a similar reason. Why, then, would the plumbob not be similarly displaced?

And if the plumbob is displaced because of its inertia, it does not indicate true up/down, and so cannot be used as a valid reference for measuring the horizontal displacement of your cannonball. Without a valid reference, the measurement of displacement is invalid, and so the results of the experiment are invalid.
Quote
How can flat earth science explain this repeatable and observable phenomenon, the easterly direction of freefalling spheres from great heights?

I don't think it really happens. I think the cannonball drops straight down.

The arc of a released object
« Reply #14 on: October 17, 2006, 01:15:06 PM »
Quote from: "Unimportant"
Quote from: "Slorrin"
These are also questions not relevant to the original one.

They are relevent to whether this experiment is valid.

Would the plumbob not be deflected in the same direction as you claim the cannonball is displaced?



Not unless it was falling.  If it was sitting still, why would it be displaced?

The force at work here is that the cannonball, travelling at a faster rate of speed being further from the earths' axis, is falling to a frame of reference which is travelling at a VERY slightly different rate of speed.

The plum bob is not travelling.  It's sitting still, in regards to it's reference.

Why would you assume it would be displaced?  It would sit still.

Do experiments with plum bobs and squares, and ou will find that that is correct, that a plum bob works.  It is in motion at the same speed as the surface f the earth.

I have found the problem with a lot of FE thinking is scale.  I read the book earth: not round and it uses TINY distances.. a couple of miles, for it's experiments.  Not thousands of miles.  

In this case, the cannonball released from the tower of pisa might land half a foot further than one dropped from the ground.

we're not talking about jet engine forces here.  We're talking about minor force differentials which are observable.


Quote from: "Unimportant"

And if the plumbob is displaced because of its inertia, it does not indicate true up/down, and so cannot be used as a valid reference for measuring the horizontal displacement of your cannonball.


That is why you would not measure your angles while walking, but while standing still.  An object at rest will remain at rest unless acted upon by an outside force.

The results of the xperiment are valid, but your understanding of the inertial forces is incomplete or flawed.

If you leave a ball at rest, it stays at rest.  Right?  But on the surface of the earth, the ball is travelling at the same speed as the surface itself.  SO relative speed is zero.  

IF you move from an altitude where the relative speed is higher SLIGHTLY because the rate of ration is the same, to a lower altitude, the relative speed of the object from high altitide, which wants to fall straight down, but continue it's forward motion at the speed of the earth's rotation (the speed it was moving at while it was being held "still" at the top of the tower).  

Since that speed is very very very very slightly different, straight down is not how it appears to fall.  It appears to fall forward.

Now, a plum bob, at one rate of travel, identical to the earth's surface, would not be displaced, at rest.

Nomatter where you hold the plum bob, it will accelerate or decerlated to meet the speed of rotation, until it sits still.  Once it sits still, it is at the same rate of speed.  It is therefor, referrentially, still.  

It would not move forward to the east.  There is no reason to believe because a cannon ball moving from high altitude to low altitude shows a slighty eastward trajectory, that a still plum line, not falling, nor moving altitude, would also have an eastward bend.

Indeed, as you say, if you walk, it trails behind you, whereas in this case, the ball moves AHEAD.  Cearly what you witness when you begin motion is the inertial tendency to remain at rest.  Whereas when it moves forward (The cannon ball) you witness it's tendency to remain in motion in a straight line unless acted upon by another force (in this case, hypothetically, gravity, or, the acceleration of the earthdisk upward.)

SO again, why would the plumbob be displaced?



Quote
How can flat earth science explain this repeatable and observable phenomenon, the easterly direction of freefalling spheres from great heights?

I don't think it really happens. I think the cannonball drops straight down.[/quote]

The arc of a released object
« Reply #15 on: October 17, 2006, 01:26:40 PM »
Quote from: "Unimportant"

Quote
How can flat earth science explain this repeatable and observable phenomenon, the easterly direction of freefalling spheres from great heights?

I don't think it really happens. I think the cannonball drops straight down.



That is not an answer.  What you think happens doens' tmatter.  What is a demonstrable phenomenon matters.  This happens, whether you want to believe it or not.

The cannonball moves eastward.

It's a fact.

I want an explanation for what causes that to happen in flat earth science.  WIthout spurious claims as to whether or not a plum bob never shows a right angle.  It obviously does show a right angle, if it created the foundations for buildings that lasted millenia, or else would they not be slanted and leaning?


I think the flaw is in how you conceptualize inertia.  For the plum bob to lean eastward, that would mean all spheres would roll eastward naturally.  Things don't just MOVE to the east without force.


If you shorten the arm of a ballista, it will throw less far.  WHy?
IF you lengthen the arm, it will throw further.  WHy?

It's because the rate of the drop of the load is similar in both cases, changing only a little with the addition of more wood and the distance of the load from the axel.  The motion is circular on a ballista.

The further from the axel, the faster the load travels.  ANd further extensions were made with the use of ropes, slings...

infact, that's how the sling works, as a weapon.. adding length to your arm so that your throw';s force can be increased by the increased speed at the further distance.

It's why if you have smaller tires on your bike, but pedal at the same rate, you go much slower.

Its the basis of differential pulleys, where two pulleys of different diameters are mounted together on the same axel, and as the pulleys make complete rotations at the same rate, one draws more cord, and the smaller one less cord, as it draws cord equal to the diameter of the wheel.

So whether you THINK the ball drops straight down, it doesn't drpo straight down.  IF you can gain access to some high towers with open windows, you can easily test that theory.

Immaone where we'd be if aristotle had tested his physical theories, rather than just conclude them by logic...

The arc of a released object
« Reply #16 on: October 17, 2006, 01:34:14 PM »
Quote from: "Slorrin"
Quote from: "Unimportant"

Quote
How can flat earth science explain this repeatable and observable phenomenon, the easterly direction of freefalling spheres from great heights?

I don't think it really happens. I think the cannonball drops straight down.



That is not an answer.  What you think happens doens' tmatter.  What is a demonstrable phenomenon matters.  This happens, whether you want to believe it or not.

The cannonball moves eastward.

It's a fact.

There's two parts to this sort of thing.

1) Prove something happens.

2) Explain why said phenomenon is evidence of a round earth model over a flat one.

You've done 2, but not 1.

Quote
I think the flaw is in how you conceptualize inertia.  For the plum bob to lean eastward, that would mean all spheres would roll eastward naturally.  Things don't just MOVE to the east without force.

No, I misunderstood what you were claiming. I knew the plumbob wouldn't deflect, but the reasoning you gave for the cannonball deflecting made it seem like you believed the force at work would be one that similarly causes the plumbob to deflect.

After reading your posts, I see you are claiming an entirely different reason for the displacement than I originally assumed. Now that we're on the same page, you've succeeded in in 2).

Good luck with 1). You can start by doing the calculations to give specific values for displacement from a certain height. Then, if someone is so inclined, they can go out and try and verify your calculations. Until then, I don't really have a reason to believe the cannonball falls anywhere but straight down.

The arc of a released object
« Reply #17 on: October 17, 2006, 01:46:47 PM »
Well, i'm glad we're on the same page.


Let's take it as read, currently that the phenomenon exists.  Can an explanation be offered for how it would exist in an FE universe?

Assuming it exists.  I can't prove it exists on a message bard without referring to something which will just be called into question, the testing ethosd, aparatus, etc..

The arc of a released object
« Reply #18 on: October 18, 2006, 05:24:49 PM »
OK, grab a pencil;
 the plumb bob is at inertial stasis in reference to a ground plane that is level and represents a vertical axis line perpendicular to it; (X)
 From this centerpoint (a) of measurement X 0, Y 0, Z 0: where X is altitude, Y is N/S and Z is E/W; locate a line segment a_b along X.
 locate 2  line segments  c_d and e_f = a_b equidistant from and parallel to a_b on Z axis.  
Locate a line segment d_b_f =  2(a_b)  parallel to Z axis line segment c_a_e  centred on X.  
We now have 2 equilateral rectangles(ref frame c,e,f,d) exhibiting X + and Z +and - and square to point (a); X 0,Y 0,Z 0. (in the analogy this is 3  ideal buildings each an ideal E/W block apart?) w/ vertex points c and e designated as target points and vertex points d and f designated as release points for equal spherical masses; objects 1 and 2 (canonballs).
 Ref frame c,e,f,d is assumed to be structurally congruent with and referentially static in any parallel Z axial plane as are 1 and 2 prior to release    
  We are looking for any Z axis deviation(n) from given X oriented d_c and/or f_e in moment path(m) of either object expressed as m1 or m2 = a_c +/- n and/or a_e+/- n.
repeated measurement under ideal, identical conditions wil, - a_c or a-e, net 0 and/ or  +/- (n).
Any Z deviation indicates action on inertial stasis of objects 1/ 2 and/or ref frame c,e,f,d.
0 = no Z deviation. a_c +n/a_e -n = Z- deviation; W. a_c -n/a_e +n = Z+ deviation; E.
 In RE model The claim is that: Z+ deviation; E is consistantly and identically indicated in actual strike points of both objects 1 and 2 being measured as a_c -n and a,e +n respectively describing path (m)as an arc of X -, Z + , Y 0.
So, 1 and 2 are assumed to have gained momentum along Z relative to ref frame c,e,f,d. imparted due to increased structural velocity of d_b_f relative to c_a_e at release. this is accounted for by viewing a_b, c_d and e_f as segments of non parallel lines convergent at point C along X at [(a) - R];approx 6350km. thus d_b_f > c_a_e.
From this an assigned constant momentum is assumed along Z from W to E and designated as C axial rotation along  Z at Y 0,X 0 as mapped to an assumed sphere.    

on a FE disk: even given an analogous rotation along any correpondent Z w/C at N pole; ref frame c,e,f,d will always exhibit Y+;N deflection/curviture  through c_a_e and d_b_f while remaining congruent. Objects 1 and 2 should describe (m) as an arc w/ Z+(n);E and Y-(n);S deflection  

If no FE rotation what about Z+ deflection?

If there is no Y-;S  deflection in the arc (m)? It becomes difficult to account for, unless the cannonballs are also interacting w/ the EM field. non mag?

Or, in a vacuum, do 1 and 2 fall straight down;d_c and f_e; w/ 0 deflection Z or Y ??      
 
So, I need 3 ideal tall buildings, a few buddies, some suveying equipment,  say, 30 bowling balls and a monster dome w/ vacuum pump. Plus currency to pay for it all,  bribe Security, etc; Piece o' cake.
 Watch for us on on the News Feeds.

Does this experiment imply that cannonballs might travel at greater velocity nearer the IceWall than at the North Pole or Equator?
 believe that; the Earth is flat until such time as I stand within the Space Station and personally see that it is a Globe.
or that the Earth is a sphere until such time as I stand upon the Icewall and personally see that it is a Flat Disk.

The arc of a released object
« Reply #19 on: October 18, 2006, 09:40:18 PM »
Quote
I don't think it really happens. I think the cannonball drops straight down


Why don't you think it would happen?
atttttttup was right when he said joseph bloom is right, The Engineer is a douchebag.

The arc of a released object
« Reply #20 on: October 18, 2006, 10:11:04 PM »
Because it conflicts with my intuitive sense of what happens to objects dropped from height. My intuition tells me they fall straight down, and I have no reason to believe otherwise.

The arc of a released object
« Reply #21 on: October 18, 2006, 10:30:33 PM »
Quote from: "Unimportant"
Because it conflicts with my intuitive sense of what happens to objects dropped from height. My intuition tells me they fall straight down, and I have no reason to believe otherwise.


There you go.
atttttttup was right when he said joseph bloom is right, The Engineer is a douchebag.

The arc of a released object
« Reply #22 on: October 19, 2006, 12:20:06 AM »
Quote from: "phaseshifter"
There you go.

There you go.

Wow! You know, I have to admit, when I saw you do it, it just looked stupid and pointless. But now that I've done it myself, I see that it's really very fun!

Thanks, phaseshifter!

The arc of a released object
« Reply #23 on: October 19, 2006, 11:30:50 AM »
Quote from: "Slorrin"


The plum bob is not travelling.  It's sitting still, in regards to it's reference.

Why would you assume it would be displaced?  It would sit still.



Not on a spherical earth.
The Plum bob would be displaced exactly the same way the bowling ball would be.

If I stand at the top of the tower, and hold the end of the string and the bob, they are traveling at the same speed.  Now I drop the bob.  It should have the same speed I do, but since I am at a greater distance above the center of rotation, If the bob maintains my speed it should seem to progress ahead of me.  The line provides tension, making the weight arc.  It's "excessive speed" is countered by being pulled up the arc ever so slightly.

However in the miniscule distances being described (the height of a tower?)

The radius of the earth is about 6400 KM.  The rotational speed of the earth is about .5km per second.  The Sears tower is 443meters tall.

So you are talking about the difference of 6400.443/6400 * .5KM/sec * over about 9 seconds for a ball to fall from a .443 KM tower. so the surface has traveled about 4.5 km while the top of the tower has gone 4.5003 km.  So if you could remove all other factors, and if you had a perfect way to measure what straight up is, the ball would hit about a foot from veritical.

But a plumb line would to, and a spirit level would also be affected.  In fact since you are in an accelerating frame work (acceleration is change of speed or direction, and by being on the surface of a spinning globe you are always under acceleration i.e. changeing direction)  you would be hard pressed to find a way to measure "true" verticle.  Even the building would probably not be built on true verticle because the forces it has to resist are in the accelerating frame.

The arc of a released object
« Reply #24 on: October 19, 2006, 02:11:01 PM »
What he said.

The arc of a released object
« Reply #25 on: October 19, 2006, 07:37:48 PM »
Quote from: "Mythix Profit"

Does this experiment imply that cannonballs might travel at greater velocity nearer the IceWall than at the North Pole or Equator?


Interesting though...If there was a clear way to establish true verticle, then the dropping ball experiment would easily establish the round or flat earth, since in the FE, the velocity of the earth increases the further you are form the pole, but in the RE it decreases with distance from the equator.

Actually in thinking of how to establish verticle (based off astonomical observation), I realised the easiest proff of the RE...The Southern Cross.

I'll have to search and see if anyone has tried that argument before.