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CES: Corning Gorilla Glass 2 is 0.8mm thick, withstands 121 pounds of pressure - Page 3

post #81 of 147
Quote:
Originally Posted by jragosta View Post

You have zero faith in the mainstream media, but you'll accept Wikipedia as a source?


Pound is traditionally defined as a measure of weight. Weight is force, not mass (That is, it is the downward force on an object caused by gravity). The confusion is caused by the fact that normal experience is that all of us experience 1 G acceleration due to gravity, so force and mass are considered equivalent when it comes to 'weight'.

Actually, the article looks like it has, let me count them.... 23 sources. And, many come straight from the National Bureau of Standards.

The old adage that WIkipedia is not a valid source discourse only needs to be mentioned for when you're writing your occasional middle school book report. For many fields, especially in the hard sciences, Wikipedia is actively maintained and well updated. And has faaaaaaaaaarrrrrr more articles than your flimsy old Funk & Wagnalls.
post #82 of 147
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Originally Posted by muppetry View Post

According to Wikipedia, slugs rarely exceed about 4 oz.

A land slug or a sea slug?
post #83 of 147
Quote:
Originally Posted by Shrike View Post

A land slug or a sea slug?

Land.
post #84 of 147
Quote:
Originally Posted by DrDoppio View Post

Pounds are units of mass, not force

You make me sad -_-
post #85 of 147
Quote:
Originally Posted by jd_in_sb View Post

My iPhone 4 fell 3 feet from my pocket and the screen completely shattered. My iPhone 4S screen came in contact with my keys in my pocket and now has a permanent scratch across it. In real-world conditions Gorilla Glass is extremely fragile. All these claims of strength are a bunch of B.S.

That doesn't mean much. My phone dropped from six feet and is fine. A lot depends on the angle of impact, and what the phone contacts on impact. Here is a pretty through test of the phone.
post #86 of 147
Quote:
Originally Posted by freelander51 View Post

the true test is putting it into your back pocket and then sitting down on a hard surface. If it does not break it's good...

Even better, will it blend.
post #87 of 147
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Originally Posted by AbsoluteDesignz View Post

You make me sad -_-

As a non imperial units user, I thought that pound was only a mass unit. I quick internet search showed me that there are two pound units...

As a reminder to anyone, mass and weight are two distinct concepts in physics. As in theory inertial mass and gravitational mass are two distinct things.
post #88 of 147
Quote:
Originally Posted by ClemyNX View Post

As a non imperial units user, I thought that pound was only a mass unit. I quick internet search showed me that there are two pound units...

As a reminder to anyone, mass and weight are two distinct concepts in physics. As in theory inertial mass and gravitational mass are two distinct things.

Not to continue the pointless quibbling over units, but the practical distinction is really between mass and force. What we commonly call "weight" is the force due to gravity here on the surface of the Earth. Only physics instructors (and other pedants :-) ever really care about the distinction between mass and weight -- in the likely event you're confined to the Earth's surface, they're practically identical. Force as a general concept, on the other hand, is easily distinguished from mass.
post #89 of 147
Quote:
Originally Posted by ClemyNX View Post

On Earth, the weight of 1 kilogram (Kg) is 9,8 Newtons (N).

That depends where you are on the Earth. For example at the equator, on the top of a mountain you would weigh slightly less due to centrifugal force being greater than, let's say, The North Pole. Also the Earth's gravitational lines of force are not an exact vector pointing to the center of the Earth either. That also varies depending on several factors including latitude, altitude, the time of year and the position of the moon. I mean, we could be talking nearly several grams of difference if you want to get technical about it.

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post #90 of 147
http://www.wsanford.com/~wsanford/gr...ewton_Ver2.txt


Bonus question: How fast much you throw a Fig Newton cookie to achieve one Newton?

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post #91 of 147
Quote:
Originally Posted by habi View Post

I must say you have a grip of steal if you have never dropped it on any hard surface. Now with iphone 4 and 4S its about 70% easier to crack the screen glass or the back glass than 3G/3GS was because of the design... No I havent dropped my iphone 4 without my silicone on but those that buy it as their first iphone maybe shocked if their not told about the fact.

The first iPhone I bought I dropped it face down from about 18 inches while getting out of my car. It hit a rock and shattered the whole face. I had only had the phone 3 days. The Apple employee felt bad for me and replaced it for free (I buy a lot of apple products from the same apple store). It occured to me a case wouldn't have prevented that damage. I don't use a case because it helps remind me to take care of the phone. The only people I know that have dropped their phone did it while opening a door or trying to hold something else in the same hand. I never do that. I always put the phone in my pocket with the screen facing in and I never put other objects in the same pocket. Follow those three rules and you won't need a case and you'll never scratch or drop your phone.
post #92 of 147
Quote:
Originally Posted by jragosta View Post

And you'd be wrong.

By definition, an item will NOT scratch something above it on the Mohs scale.
http://chemistry.about.com/od/geoche.../mohsscale.htm


This is complicated slightly by the fact that hardness can vary with the direction that you are scratching. For example, a mineral might be a 6.5 in one direction, but a 7.0 in another direction. In addition, most materials have impurities, so hardness can vary throughout a sample.



Wrong there, too.

First water does not have a Mohs hardness. Since it's impossible to 'scratch' water, Mohs hardness is meaningless for a liquid.

Second, water jet cutting does not work by 'scratching' the surface, so it is not really relevant to Mohs hardness. And for cutting steel, an abrasive is usually added, anyway.

You may be right, which is why I paused to think. However, the link you provided doesn't answer the question. It only discusses that harder materials scratch softer materials. We all know that. The question is whether a softer material can scratch a harder material. For example, can a sharp soft material scratch the planar surface of a harder material. Clearly at some amount of force it can. At the point of atom bombardment, soft materials can clearly scratch a harder material. Maybe that isn't considered a scratch?? Fair enough. I'm just trying to figure out why I have a couple of micro scratches in the back of my iphone. If a softer material really can't scratch a harder material (despite being sharp and a force being applied) then it must be a small rock or piece of sand in my pocket that made a scratch.

LOL with regard to your enlightenment about water not having a mohs hardness. No Shit.
post #93 of 147
Quote:
Originally Posted by muppetry View Post

That was my experience until one day I found a significant scratch on the front of my IP4. Never did figure out what it might have come into contact with to do that. Unless it was my BlackBerry...

I sometimes use our ottoman (footstool) as a desk. I will then sometimes charge my iPhone via the usb port on the unibody macbook to charge my iPhone. With kids and pets running around sometimes the computer gets bumped or nudged and it slides up and over the phones screen. I believe that the anodized coating on the Macbook or the Apple wireless keyboard is hard enough to scratch the gorilla glass.
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post #94 of 147
Quote:
Originally Posted by mstone View Post

That depends where you are on the Earth. For example at the equator, on the top of a mountain you would weigh slightly less due to centrifugal force being greater than, let's say, The North Pole. Also the Earth's gravitational lines of force are not an exact vector pointing to the center of the Earth either. That also varies depending on several factors including latitude, altitude, the time of year and the position of the moon. I mean, we could be talking nearly several grams of difference if you want to get technical about it.

Not to forget that it's 9.81 m/s^2 in the acceleration assuming freefall.
post #95 of 147
Quote:
Originally Posted by mstone View Post

That depends where you are on the Earth. For example at the equator, on the top of a mountain you would weigh slightly less due to centrifugal force being greater than, let's say, The North Pole. Also the Earth's gravitational lines of force are not an exact vector pointing to the center of the Earth either. That also varies depending on several factors including latitude, altitude, the time of year and the position of the moon. I mean, we could be talking nearly several grams of difference if you want to get technical about it.

Yep.

I'm thinking of an infomercial advertising a guaranteed, foolproof weight loss method. For $10,000, I'll send them a ticket to the equator.
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post #96 of 147
This CES feels weak. Too many companies playing catchup to Apple. The most interesting thing is an advancement on fraking glass. Glass!

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post #97 of 147
Quote:
Originally Posted by SolipsismX View Post

This CES feels weak. Too many companies playing catchup to Apple. The most interesting thing is an advancement on fraking glass. Glass!

I thought Intel actually getting closer to shipping a viable (not necessarily competitive) x86 SoC for smartphones and tablets is some pretty big news. And they got a couple of big name suckers (Moto and Lenovo) to make some phones for them, running an x86 version of Android.

If they can compete, it could be a very interesting thing. The half-life on this thing has been a long long time coming, and it's still at least 6 months away.
post #98 of 147
Quote:
Originally Posted by Shrike View Post

I thought Intel actually getting closer to shipping a viable (not necessarily competitive) x86 SoC for smartphones and tablets is some pretty big news. And they got a couple of big name suckers (Moto and Lenovo) to make some phones for them, running an x86 version of Android.

If they can compete, it could be a very interesting thing. The half-life on this thing has been a long long time coming, and it's still at least 6 months away.

I'm not betting on that until we start seeing some real devices. Getting some partners is okay but if Intel really has made their architecture mobile we'll see vendors move from ARM willingly and in droves.

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post #99 of 147
They need to get Haswell out. They need to get Haswell out YESTERDAY. The touted massive battery life improvement will be the clincher in creating x86 tablets if that's their bag.

I'm rooting for Apple's choice, but it's silly to think that ARM designs will have any reason to get better if no one can compete against them. I'm rooting for Intel to compete as they have always done.
post #100 of 147
Quote:
Originally Posted by muppetry View Post

Land.

African or European?
post #101 of 147
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Originally Posted by dcsimages View Post

African or European?

Huh? I... I don't know that.
post #102 of 147
Quote:
Originally Posted by muppetry View Post

Huh? I... I don't know that.

It's a joke from Monty Python. I was making an oblique reference to it with sea or land slug. African and European is just a dead give away.

Since you don't know, you're going to be pushed off the cliff now.
post #103 of 147
Quote:
Originally Posted by Shrike View Post

It's a joke from Monty Python.

He's providing the in-movie response.

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Since you don't know, you're going to be pushed off the cliff now.

Indeed.
post #104 of 147
Quote:
Originally Posted by Shrike View Post

It's a joke from Monty Python. I was making an oblique reference to it with sea or land slug. African and European is just a dead give away.

Since you don't know, you're going to be pushed off the cliff now.

Yes - sorry - I should have been clearer, but I assumed that you were expecting the Bridgekeeper's reply.
post #105 of 147
Some additional info in this BBC article: http://www.bbc.co.uk/news/technology-16523817

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post #106 of 147
Quote:
Originally Posted by SolipsismX View Post

Some additional info in this BBC article: http://www.bbc.co.uk/news/technology-16523817

Nice article, thanks. Their stock has halved this past year; perhaps time to invest now ($14.32) GLW, NYSE.
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post #107 of 147
Quote:
Originally Posted by ClemyNX View Post

As a non imperial units user, I thought that pound was only a mass unit. I quick internet search showed me that there are two pound units...

As a reminder to anyone, mass and weight are two distinct concepts in physics. As in theory inertial mass and gravitational mass are two distinct things.

Well let's confuse it more and recognize that a pound is also a unit of currency! I don't think anyone disputes the separate natures of mass and force. There was just a lot of digging in over what pound refers to exactly, which is both of those and more. All contextually dependent.
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post #108 of 147
Quote:
Originally Posted by ash471 View Post

I'm just trying to figure out why I have a couple of micro scratches in the back of my iphone. If a softer material really can't scratch a harder material (despite being sharp and a force being applied) then it must be a small rock or piece of sand in my pocket that made a scratch.

It's the bold underlined part, or some similarly hard material particle that made it into your pocket or onto some other item in your pocket which is itself softer.


Quote:
Originally Posted by bigdaddyp View Post

I sometimes use our ottoman (footstool) as a desk. I will then sometimes charge my iPhone via the usb port on the unibody macbook to charge my iPhone. With kids and pets running around sometimes the computer gets bumped or nudged and it slides up and over the phones screen. I believe that the anodized coating on the Macbook or the Apple wireless keyboard is hard enough to scratch the gorilla glass.

Or like the above, there was some sand residue on the ottoman, which somehow got between the iPhone and the MacBook case. Now you have the proverbial rock between two hard places. The particles that can do the scratching can be really small, small enough to feel like smooth powder to a finger, if they are pinched down by another hard surface.
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post #109 of 147
Quote:
Originally Posted by mstone View Post

That depends where you are on the Earth. For example at the equator, on the top of a mountain you would weigh slightly less due to centrifugal force being greater than, let's say, The North Pole. Also the Earth's gravitational lines of force are not an exact vector pointing to the center of the Earth either. That also varies depending on several factors including latitude, altitude, the time of year and the position of the moon. I mean, we could be talking nearly several grams of difference if you want to get technical about it.

The centrifugal [centripetal] force has nothing to do with not at all, that would actually be trying to throw you off the mountain and would be higher on a high mountain. The gravity difference is simply that you would be farther from the Earth's center of mass, so your r^2 term is larger, making the force proportionally smaller. Latitude has the same effect because the Earth is an oblate spheroid bulging towards the equator, not a true sphere, so both distance and mass relative positions are affected. Other mass effecting properties of the Earth not being a homogenous constant density sphere hold true at the same time.

And the differences on a mountain would be in thousands to hundredths of a gram, weight-wise. Even in orbit a body still has a lot of "weight", it just isn't being pushed back-on by a fixed surface. The orbiting body is just traveling so fast sideways it is roughly canceling out the falling part of the motion and you never get to a place to stand or go splat.
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post #110 of 147
Quote:
Originally Posted by Hiro View Post

The centrifugal force has nothing to do with not at all, that would actually be trying to throw you off the mountain and would be higher on a high mountain. The gravity difference is simply that you would be farther from the Earth's center of mass, so your r^2 term is larger, making the force proportionally smaller. Latitude has the same effect because the Earth is an oblate spheroid bulging towards the equator, not a true sphere, so both distance and mass relative positions are affected. Other mass effecting properties of the Earth not being a homogenous constant density sphere hold true at the same time.

And the differences on a mountain would be in thousands to hundredths of a gram, weight-wise. Even in orbit a body still has a lot of "weight", it just isn't being pushed back-on by a fixed surface. The orbiting body is just traveling so fast sideways it is roughly canceling out the falling part of the motion and you never get to a place to stand or go splat.

He meant that the earth's rotation affects gravity as a function of latitude, not altitude. At the equator, you are both further from the earth's center of mass and rotating around it. Those two actually produce roughly equal reductions in weight at the equator relative to the poles - of the order of 0.5%.

With altitude above the earth's surface at the equator, the effect is dominated by distance from CoM, since g varies inversely with r², whereas the centrifugal force increases only linearly with r.
post #111 of 147
Quote:
Originally Posted by muppetry View Post

He meant that the earth's rotation affects gravity as a function of latitude, not altitude. At the equator, you are both further from the earth's center of mass and rotating around it. Those two actually produce roughly equal reductions in weight at the equator relative to the poles - of the order of 0.5%.

With altitude above the earth's surface at the equator, the effect is dominated by distance from CoM, since g varies inversely with r², whereas the centrifugal force increases only linearly with r.

I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.

You can't even construct the virtual centripetal [centrifugal] force vector because the solid surface is under the person standing, not on the outside radius as it is in a centrifuge. There simply is no force trying to throw things off the surface of the Earth because the earth is spinning, so there is no reduction in gravity because of it.

Differences in gravity between Poles and equator are due to the shape of the Earth and differences in local composition density contributions.

The only thing about escaping gravity that is good about the equator is that the speed for orbit is independent of the launch point. 17K MPH is 17K MPH whether you launch from Alaska or Canaveral, but closer to the equator you are moving faster and if you launch in the same direction as rotation you don't need as much fuel to accelerate to that 17K MPH, because you start with 1K MPH at the equator.
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post #112 of 147
Quote:
Originally Posted by Hiro View Post

I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.


Maybe we should start by reading the wikipedia page on gravity.

http://en.wikipedia.org/wiki/Gravity_of_Earth

I know it is not really an authoritative source but... just saying.

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post #113 of 147
Quote:
Originally Posted by mstone View Post

Maybe we should start by reading the wikipedia page on gravity.

http://en.wikipedia.org/wiki/Gravity_of_Earth

I know it is not really an authoritative source but... just saying.

It's a fine source.

Quote:
Originally Posted by Hiro View Post

I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.

You can't even construct the virtual centripetal force vector because the solid surface is under the person standing, not on the outside radius as it is in a centrifuge. There simply is no force trying to throw things off the surface of the Earth because the earth is spinning, so there is no reduction in gravity because of it.

Differences in gravity between Poles and equator are due to the shape of the Earth and differences in local composition density contributions.

Let me give you a question to consider. I assume that you would agree that on the equator, one is moving (relative to the center of mass, M, of the earth) in a circular path, of radius equal to the equatorial radius of the earth, R, and a period of one day. In which case, would you agree that your motion includes a radial acceleration, a, (normal to your instantaneous velocity vector) given by

a = - ω²R , [1]

and that to maintain that acceleration (circular motion) requires a centripetal force, F, given by

F = - ²R, [2]

where m is your mass and ω your angular velocity?

Hopefully you would agree, because that is simple Newtonian physics. That centripetal force is, of course, provided by a fraction of your gravitational attraction to the earth, and your apparent weight, mg (the residual reaction between you and the surface you are standing on), is reduced by that amount. Or more rigorously, if G is the gravitational constant and g is the acceleration due to gravity in the local coordinate system (your initial acceleration if you jump off a building), for mechanical equilibrium between yourself and the surface of the earth:

mg = GMm/R² - ²R , [3]

While there actually is no such thing as a centrifugal force - it is just Newton's 3rd law reaction to a centripetal force - there is indeed a centripetal force acting in this system.

So, I assume that you are drawing the distinction between pure gravitational attraction and actual acceleration due to gravity in a dynamic system. If the discussion is static gravitational attraction then you are correct, but when discussing actual acceleration due to gravity (i.e. the acceleration of an object in free fall in the local coordinate system), the angular velocity, ω, of the coordinate system has the effect described, leading to an acceleration given by dividing [3] though by your mass, m,

g = GM/r² - ω²r . [4]

The second term on the RHS is the effect of the earth's rotation. Does that explicit distinction reconcile the two views?
post #114 of 147
Quote:
Originally Posted by muppetry View Post

It's a fine source.

I don't disagree with it, but it is being horribly misinterpreted in this thread. Gravity is gravity. Apparent gravity (which is the part of that page which is being referred to without proper disambiguation) is an entirely different phenomena that takes everything into account including hiccups. It isn't gravity at all, but just a resultant force of all forces and accelerations currently acting on a body. And many shorthand the name as apparent gravity because the direction is almost the same as the local true gravity vector.

Quote:
Originally Posted by muppetry View Post

Let me give you a question to consider. I assume that you would agree that on the equator, one is moving (relative to the center of mass, M, of the earth) in a circular path, of radius equal to the equatorial radius of the earth, R, and a period of one day. In which case, would you agree that your motion includes a radial acceleration, a, ...

No, I wouldn't agree with that. Bad assumptions. You are trying to measure from the wrong frame and wrong coordinate system to make that direct comparison. When the frame of reference is the center of rotation of the Earth there is no radial acceleration of an object on it's surface because the Earth is for all intent and purposes rotating at constant velocity (on a scale of billions of years). And an object on the surface remains in the same absolute & relative position on the surface. When you use the correct reference frame and a 3D polar coordinate system that is compatible with the 3D shape of the Earth there is no need for virtual forces to balance out weird interactions of a 2D projection of a 3D system. In the polar fixed system the answer are dead-simple vector-linear relationships to the cg of the Earth which follow the force over squared-distance rule.

The rest of the apparent gravity portion of the post was superfluous because it depended on the entering assumption which I completely disagree with. And it's a good thing I do because there is no skyhook to grab us and provide the centripetal force you want to find. If you want to measure from a non center of the earth frame and talk about earth surface gravity (not the fake apparent gravity) you have a hell of a lot more math to do to keep everything correct.

Quote:
Originally Posted by muppetry View Post

So, I assume that you are drawing the distinction between pure gravitational attraction and actual acceleration due to gravity in a dynamic system.

Again I'll stop you right here. Gravity is gravity, there is no difference between static and dynamic gravity. The force and acceleration it imparts is the same whether you are talking static or dynamic. You are confusing the fact that when you measure gravitational force from two different reference frames the numbers will look different, because they are relative to some other dynamic thing, but the actual gravitational force and acceleration vector will be in the same direction and have the same magnitude. All because the vectors themselves are absolutes, independent of the frame. You just perceive the projections of these absolute vectors onto the alternate reference frames to be different from the relatively moving frame reference you are measuring from.

It is dangerous and incorrect to compare values measured in one frame to values measured in a different frame without applying a correction factor taking into account the difference in relative motion between the frames. Sometimes that is straightforward, sometimes it is an all-out-complete-cluster-mess. And the all-out-complete-cluster-mess is common enough that is is just a damn sight easier to remember to only compare inter-frame rather than intra frame.


Just keep it simple. Gravity is gravity, the attractive force between objects with mass. If you want to combine gravitational vectors with other force vectors, that works because D'Lambert proved so, but the resultant vector isn't gravity, it is just the resultant vector.
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post #115 of 147
Quote:
Originally Posted by Hiro View Post

It isn't gravity at all, but just a resultant force of all forces and accelerations currently acting on a body.

That is exactly what the original thread is about. Since the gorilla glass has very little magnetic properties the force it can withstand is not affected by gravitation either on Earth or in space. The discussion was slightly diverted by the comment that 1 kg equals 9.8 N which is such a rounded average that it would be true even if a kg only had 995 g or as much as 1005 g which by definition is completely unscientific. I apologize for being overly terse and unnecessarily technical but when you examine physical properties from an analytical perspective you have to consider every possible implication that the laws of physics dictate.

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post #116 of 147
Quote:
Originally Posted by Hiro View Post

When the frame of reference is the center of rotation of the Earth there is no radial acceleration of an object on it's surface because the Earth is for all intent and purposes rotating at constant velocity (on a scale of billions of years).

OK - well I guess we can stop right there. Your argument is that an object moving in a circle (represented by the motion at the earth's surface due to its rotation about its center of mass) has no centripetal acceleration because it has constant angular velocity? You really want to stick with that? Because if so then you are woefully ignorant of basic physics. That you choose to demonstrate that ignorance by arguing with a physicist is kind of unfortunate, because I can't let that kind of error go uncorrected. Actually "arguing with" is a bit of a stretch. You are simply contradicting me.

Centripetal acceleration is just ω²r - it depends on the square of the angular velocity, not the rate of change of angular velocity. All objects in constant circular motion have a constant and non-zero radial acceleration by definition. Since the centripetal (radial) acceleration vector is normal to the (tangential) velocity vector it results only in a change in direction (that which causes circular motion), with no change in the magnitude of the velocity vector (speed). Perhaps that is what is confusing you. They are separate.

And by the way - when I refer to "coordinate system" I don't mean projection or anything else related to maps - it is entirely shape independent and I'm referring to the frame of reference in Euclidean space.

In any case, attempting to dismiss the analysis by vague mis-application the laws of motion will not work. If you would like to take a more rigorous stab at proving your assertion then be my guest. You will fail though.
post #117 of 147
So we're arguing the very clear definition of gravity vs the effects of gravity of a given planetary body acting upon an object when accounting for effects that would negate the perceived effects of gravity at a given location?

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post #118 of 147
Quote:
Originally Posted by SolipsismX View Post

So we're arguing the very clear definition of gravity vs the effects of gravity of a given planetary body acting upon an object when accounting for effects that would negate the perceived effects of gravity at a given point?

Well I thought that was exactly what we were arguing about until our friend started with the puesdo-scientific stuff and then massacred some of the basic laws of physics. Then it stopped just being two different viewpoints on the same thing and became a physics lesson.
post #119 of 147
Quote:
Originally Posted by majjo View Post

Oh god, this thread is still going?

No, they both agree on that. What they're arguing is the existance of the centripetal force due to earth's rotation.

I know - it's embarrassing that this has gone on for so long, but it's hard to drop. And just to clarify, the centripetal force is not due to the earth's rotation, it is required to keep an object on the equator moving in a circular path, and is provided by a (small) part of the gravitational attraction between the object and the earth. The remainder of the gravitational attraction is what we perceive and measure as weight, and is where the discussion started - with someone or other correctly mentioning that the rotation of the earth reduces weight at the equator relative to at the poles.

The argument began when another poster disputed that, but I thought at first that he was really just trying to point out, as Soli summarized, that the magnitude of the gravitational field vector at a point in space is independent of the frame of reference of the observer, provided that frame of reference is steady and non-acceleratory. That, too, is correct, but as an observer standing on the equator, that condition is not satisfied because your frame of reference is accelerating (centripetally) and the measured gravitational attraction (weight) is reduced accordingly. However, it then turned out that he was actually wrongly disputing the very existence of centripetal acceleration in steady circular motion.

Or possibly he just responded in too much of a hurry and didn't really mean to say that. If he replies again we may find out.
post #120 of 147
Quote:
Originally Posted by majjo View Post

Oh god, this thread is still going?

I love this forum because of the very diverse and in-depth conversations(arguments) that often occur. I am a member of many forums yet this only happens on AI's forum. It's great! I certainly learned something about gravity from this thread.

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"The real haunted empire?  It's the New York Times." ~SockRolid

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