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Online Nominal Animal

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Re: Physics Question - ma = mg
« Reply #100 on: June 28, 2021, 04:26:20 pm »
You can measure mass (as opposed to weight) using angular momentum and torsion springs.

By 1785, those were already so sensitive that Charles-Augustin de Coulomb could use one to measure electrostatic forces, and came up with Coulomb's law.

By 1798, Henry Cavendish used one to directly measure the gravitational attraction between two masses in a laboratory experiment, the famous Cavendish experiment, although geologist John Michell designed the experiment (Coulomb following his plans), sometime before 1783, but alas, kicked the bucket too early. (Michell was 69 when he died, though.)

With commercially available load cells, a turbo pump for pumping a good vacuum, some glass jars and such, you can definitely measure for example the gravitational attraction between say two marbles, in a freshman physics experiment.  The most expensive thing is getting precise mechanics nowadays.  A couple of hundred years ago there were watchmakers and others happy to help produce interesting scientific apparatuses, but now, you're better off with old mechanisms and new sensors!

1.  Please stop calling g a physical constant, in the sense that G (universal gravitational constant) is a constant.
Hey, I blame the language!  We need a word describing a variable that is not being varied in this particular context.

For differential calculus, we use "derivative" and "partial derivative"; so would "partial constant" work?



Yesterday, I spent a couple of hours admiring the concept and word geodesic.

Let me waffle on a bit once again, but this time, as a hopefully entertaining but informative musing on how precise, exact terms make life interesting, and mushy overloaded ones a misery.  (I am thinking of making a T-shirt with "Why don't you just use pastel tones to reduce gravity?, too.)

As a background, geodesic is when you extend the concept of "straight line distance between two points" to surfaces; and from its name, one of the most used one is the "straight line distance" between two points on the surface of a sphere or geoid, like the Earth.  On a perfect sphere, that distance is the shorter arc of the great circle – a circle passing through those two points, with its center at the center of the sphere.

What happens, if you ask your Californian friend what the distance between LA and Las Vegas is?
My bet is that the most common answer is something like "Oh, about four hours, at this time of day."
Bitch, I asked you distance, and you gave a time interval.  What sort of mindfuck is this?

(If you happen to have the cultural context, you'd know that because primary transport in that region is via personal automobiles, and the physical distance is less relevant than the actual travel speeds achievable, it makes more sense to describe distances using the typical time taken to drive that distance, than anything else.  Just like we include the completely unrelated inertial term in our "average acceleration on Earth due to gravity", because we don't have any better place to put it, and keeping it there takes care of it very nicely, thank you.)

In a different cultural context, that friend might ask for clarification, say "As the crow flies, or?".
Bitch, I ain't an ornithologist or an ornithopter; how the hell would I know how corvids fly anyway?  They may do so ass first for all I care.

Now, let's get to "geodesic".

Let's say you're driving from LA to Las Vegas with your friends' kid on the back seat, thinking about normal kid stuff, like how many hydrogen bombs in the 50 to 100 kiloton range would they need to carve a canal from Los Angeles to Las Vegas.

The kid asks, "Do you know what the geodesic between Los Angeles and Las Vegas is?"
Why, you just looked that up a few days ago for this very thread, so you answer "About 368 km, but there is also an elevation difference of about 525 meters."

SEE? No confusion.  Straight, unambiguous answer unrelated to ornithology, free of oddities like trying to measure distances using time units, and so on.
The kid happily takes the added cratering depth into account, you both smile, and have a nice car drive.  Everybody wins.

If you try to be physically or geometrically correct but do not know the term, the mistakes and confusion one gets mired in are endless.
For example, let's say you ask "What is the distance between Los Angeles and Las Vegas?", I ask for clarification, for which you say "in a direct line, but ignoring the curvature of Earth and the difference in elevation".  My answer, not to be snarky but maximum helpfulness, could likely be "Either 368 km or 39707 km, depending on which way you measure it."
I know, I'd get bitch-slapped for that (if I wasn't driving, and I don't).  But I lack the context necessary to give the "straight answer"; I either have to make assumptions (which are often wrong), keep asking for clarifications, or give an answer that outlines the possible set so the asker can determine what the hell they were trying to find out in the first place.

And no, I don't really like "partial constant".  But I'm used to it, because I use const every day, and it does not mean that something is constant, just that in that particular context I (or the C code I write that uses that) promises to the compiler to not try and modify that value.  And volatile does not mean "flames or vapours are imminent, get your fire extinguisher ready!", it means "hey compiler, that value may change at any point, so make no assumptions about it okay?".  And I have to talk about stuff like "immutable string literals" to get new programmers to understand you really don't get to modify them; that if you do, you broke a promise, and the kernel gives you a segmentation fault in confusion.

But I perfectly understand the frustration with mushy and misused concepts and terminology.  If we'd have a better set, one with say posters or web pages intended for different levels of complexity/understanding, that I could point others to and say "using these" ..., I would do exactly that.  And I'd be darn happy; I'd buy it as a dozen posters and lovingly put them on my walls and give out as gifts; they'd make that much a real-world measurable difference.

But scientists and their insistence of attaching peoples surnames to things yielding undescriptive (or in the case of original discoverer being someone else, sheer misleading history-twisting self-aggrandizement) words one has to learn by rote by the dozen and use "correctly" (read differently) in every single little domain, and science becomes a Black Magic only those versed in the Old Powerful Names of Power and History and Power have the sheer single-mindedness to delve into.

I mean, I can say something like "My mother in law was being a sheer Cheney this week visiting us.  Thank Dog her Trumpy sister could not come!  Dealing with those two is like trying to watch The Kardashians and Dr. Phil at the same time.  I don't know... I think I might be getting a bit Biden in my old age or something; I've been acting like a complete Dubya." and people would just get it.

But use "mass" and "constant", and people start getting more confused and even angry... No, this is not sane or right.  Yet, it is the world we have.

So, what is one to do?  Even writing about this like I've decided to do in this thread gets me flak about wasting others time, and gets me added to new ignore lists.  I don't know.  I just pick a topic at a time, think hard about it, and especially all the ways I know and think of it can be misunderstood, and if I decide it seems possible/worthwhile/useful, construct something that hopefully helps others wade through the word salad but with new mental tools and understanding at their disposal.
« Last Edit: June 28, 2021, 04:29:30 pm by Nominal Animal »
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #101 on: June 28, 2021, 04:44:48 pm »
In technical English, a "parameter" is either a constant that can be varied, or a variable whose value is held constant during the calculation.
The popular media keep confusing "parameter" with "perimeter", the boundary around an area.

"Geodesic" is a very interesting concept.  In geometry, it is the distance between two points on the manifold in question with the shortest length.  In three-dimensional Euclidean geometry, it is merely the straight line between the two points.  On the surface of a sphere, it is the great circle (arc with center at the center of the sphere) that connects the two points.  Traditional ocean navigation plots the great circle on a globe or appropriate projection flat map, and then transcribes the points onto a piecewise-linear route on a Mercator projection, where any line segment is a line of constant compass heading, appropriate to the helmsman.  In a classic textbook on general relativity, one example is an ant crawling on the surface of an apple (not really spherical), where the claim is made that sensing only the spatial derivatives along its short body, the ant will follow a geodesic path.  That is a good parable, but I don't know if anyone has done the experiment with a real ant, or if it is literally applicable, since I don't know if the ant has a definite endpoint in its tiny mind (analogous to LA and LV).

The traditional measurement of mass is to use a gravitational balance (such as held up by Lady Justice at the courthouse) and compare the mass in question to a standard mass, defined by the King.  As a practical matter, a "steelyard" balance (q.v.) uses leverage to compare the mass in question to one or more standards.  Technically, this equates the gravitational masses on both sides of the balance, but standard theory and Roland, Baron von Eötvös tell us that gravitational mass and inertial mass are equivalent, hence can be set equal.
« Last Edit: June 28, 2021, 04:59:36 pm by TimFox »
 

Online Nominal Animal

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Re: Physics Question - ma = mg
« Reply #102 on: June 28, 2021, 08:58:53 pm »
In technical English, a "parameter" is either a constant that can be varied, or a variable whose value is held constant during the calculation.
The popular media keep confusing "parameter" with "perimeter", the boundary around an area.
That works.

Before I encountered the term eccentric angle, I consistently called \$\theta\$ "the angular parameter" in
$$\left\lbrace\begin{aligned} x &= a \cos \theta \\ y &= b \sin\theta \end{aligned}\right. \quad \iff \quad \left\lbrace\begin{aligned}
\varphi &= \arctan\left(\frac{b}{a}\tan\theta\right) \\
r &= \frac{ab}{\sqrt{(b \cos\varphi)^2 + (a\sin\varphi)^2}} \\
\end{aligned}\right. \quad \iff \quad \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where the leftmost set gives the Cartesian coordinates for the points on the axis-aligned ellipse as a function of the angular parameter, or eccentric angle \$\theta\$; the middle one is the same in polar coordinates where \$\varphi\$ is the polar angle and \$r\$ the distance to the ellipse from origin in that direction; and the right side being the familiar implicit form in Cartesian coordinates.

(Oh wait, it isn't an implicit form, because I left the one on the right side.  Whatchamacall this form then? Pre-implicit form?  Explicit form without any nekkid stuff?)

It is so easy to accidentally forget that just because you have a \$\theta\$ or \$\sin\theta\$ or \$\cos\theta\$ term, does not mean that \$\theta\$ necessarily refers to a particular angle.  Eccentric angle is illustrative, because you immediately intuitively know that "Ha! I bet it means it is not exactly the polar angle!".

I like "angular parameter" too, because for e.g. Lissajous curves (\$x = A \cos(a \theta + a_0)\$, \$y = B \sin(b \theta)\$) \$\theta\$ really is just a parameter you need to vary to sweep the entire curve.

"Geodesic" is a very interesting concept.  In geometry, it is the distance between two points on the manifold in question with the shortest length.
Yes; I didn't want to write that myself, because the concept of manifold would have to be explored first.

Case in point:  A couple of years ago, I helped a CS or SE major with a raytracer visualizing non-Euclidean space.  It was piecewise Euclidean; essentially modeled using simplices (tetrahedra) whose faces are discontinuities yielding an overall non-Euclidean geometry.

If you've seen the Cube movies, then you probably know the idea: each simplex is a "room", with arbitrary other "rooms" sharing walls in arbitrary orientations.  Aside from the discontinuity (a coordinate transform we used barycentric coordinates for, yielding a pretty useful geodesic) at the faces of each simplex, the "rooms" themselves were Euclidean.  It modeled pretty much perfectly the case where you're looking at the back of your own head, except with just simple geometric primitives.

I still don't know whether you can consider that a manifold, because of the discontinuities at the tetrahedral mesh faces.  "Piecewise Euclidean" suited me better; I don't know what they used in their report/paper/whatever.  (In Cube, the walls are always perpendicular, so there are no discontinuities of the sort the tetrahedral mesh has; that I do believe is a proper manifold.)

I also tried to avoid dealing with terms like "embedding", because defined the way it was, it is darned near impossible to give a nontrivial simulated model any specific dimensionality.  If you take a cube, and "glue" its faces together permuting their orientation, what is its dimensionality?  I think there might even be pathological cases (choices of mating face pairs and their relative orientations) where you'd need infinite number of dimensions to embed that into a higher-dimensional Euclidean space.
If you consider how the tetrahedral meshes can be constructed, it is easier to define the meshed topology in terms of its internal consistency and structure, and completely ignore any efforts to see say where the tetrahedral lattice nodes would be if you wanted to assign them Euclidean higher-dimensional exact coordinates.

I preferred to just show the tools, and leave it to mathematicians to define and help with the correct terminology.  But I know for a fact that the darned thing worked, because I've seen the rendered images and fully understand how the math works and why.  But put me among a gaggle of mathematicians, and they'll laugh me out of the room as I fumble trying to find the correct terms Words of Wisdom and Power and I Even Have This Hat You See to describe what math it is based on.

If you think about that, it is kinda sad.

Reminds me of the joke about a man who wanted to see the Circus Director.  The director was busy, so snapped at the man to tell what he could do, and to be quick about it.  The man proudly exclaimed he could emulate any bird!  The director laughed at the man, saying whistlers and ventriloquists are a dime a dozen, and to get out; he was just wasting the Director's valuable time.  Utterly dejected, the man sighed and flew away.

(The reason I helped with that, is that I once had a dream test-playing a game I purportedly designed, somewhere between Marathon and Doom, but with exactly the sort of non-Euclidean geometry as described above.  Basically, any enclosed tubular region of the game world, the world isn't described in absolute coordinates at all, but as piecewise Euclidean spaces glued together using minimal shear discontinuities.  A particular example is of a slightly sloping tunnel/stairwell upwards, where you walk "around the corner" (in a circular arc) some 700-800 degrees, without returning to your original location at all.  Not just "bigger on the inside", but no turn or bend ever goes the way you think it does. This is trivial to implement with suitably chosen data structures in almost any 3D graphics/physics engine, but at least in my dream the "uneasiness" of going one way, even coming back and going again the same way, but arriving at a completely different place you think, and not being able to find a discontinuity where the "error" or "trick" occurs (there really is none, they're within rounding error; it's just not really Euclidean, only piecewise Euclidean), was fun-kyy.  I don't even like that sort of games, although I've played a little at parties for the social aspect; I have even less interest in designing or creating such games.  I see Odd dreams, I guess.)
« Last Edit: June 28, 2021, 09:02:38 pm by Nominal Animal »
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #103 on: June 28, 2021, 09:17:57 pm »
Yes, I agree, having the context of using 'g' in a varied environment, not just at Earth surface, and simplifying to be a smooth ball, uniform density, etc.
  So then "...local acceleration conditions..." for lack of a better word.
   Sounding better and better.
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #104 on: June 28, 2021, 10:08:48 pm »
I think the better words are "g is the acceleration of gravity in a region of uniform gravitational field".
 

Online CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #105 on: June 28, 2021, 11:34:17 pm »
After following many pages of this I think I have finally realized another of the problems.  It is akin to the pound mass pound force problem. Two different things with the same name.  g has even more definitions for the same symbol.

Some practitioners use g to refer to gravitational acceleration at the surface of the earth.  It has a formal value, traceable to the defined value of G, and can be used for many calculations that don't require corrections for centripetal acceleration, mascons and the like. 

Other practitioners uses the same symbol to refer to the local corrected acceleration (somewhere on or very near the Earth) including all of the correction factors.

And still another group uses it it refer to the nominal acceleration  due to the dominant local mass, the earth, moon or mars for example.

I personally assume that the first usage is intended unless context suggests otherwise.  That works in the vast majority of cases, especially if you use the sub rule to check the context more carefully the more decimal places matter.

 The third usage is uncomfortable to me, but made more palatable by flagging and identifying with a subscript identifying the reference mass.  Something which I haven't noticed happening in this thread.
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #106 on: June 29, 2021, 06:27:23 am »
I vote for a lable of:

        'REGIONALLY STABLE VOLITILE'

to use 'g' in a C code and defined as Volitile to C compiler.
but that's a bit snarky / infantile. Hey, some classic jokes provide a sort-of structure, for framing difficult / frustrating problems. (I'd rather that irk, over a direct, energetic encounter with gravity...Oops, I meant 'rapid deceleration (w ground).
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #107 on: June 29, 2021, 01:03:02 pm »
While I agree that 'g' should not be called a constant, a few errors have taken place throughout this thread.

The whole topic is based around ma = mg. If this were a topic of C code and someone used 'g' as a variable, then 'g' would be known as a variable, however, it began (and continued) around ma = mg.

Somewhere along the way someone (or at least how I interpreted) needed to state that 'g' is the acceleration of gravity on Earth and 'G' is the Gravitational constant. I was a bit baffled why someone assumed the contributors of this thread were confused over g and G when we clearly separated the lower and upper case.

In any case, I feel some began reading the thread halfway and therefore felt 'g' was improperly used. Much like if I began reading halfway through about a programming thread involving 'g' as a variable, I may ask: why are you using the gravity on Earth as a variable.

I'll ask the philosophical question though: is anything really a "constant"?

If science is always polishing the apple per se, then every 'constant' we know will change. Pi is a constant, but it's a never ending value that hasn't been fully carried out (nor may never). G is the gravitational constant, however, it must have some error margin.

If all these numbers are constantly being worked on to be more precise, then are they truly constants? By saying (little) 'g' is a constant may not be completely incorrect.
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #108 on: June 29, 2021, 01:53:26 pm »
If you define g as the acceleration of gravity at the surface of the earth, where exactly on that surface is that to be measured?  It is slightly different at the peak of Everest, in the city of Denver, and at whatever they are calling mean sea level now.  I prefer g as a parameter measured within a region of interest where the gravitational field is uniform, whether that be on Earth or on the moon, which should be stated or made obvious in context.
The importance of g is that all objects within this region of interest experience the same value of gravitational acceleration, independent of mass.  This is different from Aristotle's theory that predates Newton.
(Of course, all of this neglects air resistance, as in freshman physics discussions.  When there is serious resistance, the force is proportional to velocity, and dependent on the object's shape.  If you drop an object far enough through a medium, the fall approaches "terminal velocity", where the resistance force equals the gravitational force and the object stops accelerating.)
 

Online CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #109 on: June 29, 2021, 02:03:56 pm »
In a practical sense g and G are constants.  By agreement of international standards bodies.  While the values may change over time, in these two cases, and over human time scales these changes will be within the error bars of the existing definition, somewhat akin to computing more digits of pi.

I'll leave the philosophical questions of whether these matter on some cosmic sense, and programming style questions to others.

As always, when doing something that has real consequences it is important to thoroughly understand the problem at hand.  A programmer who uses a standard constant library which includes g better not use that for a Mars lander.  A student taking a test needs to understand what the teacher is asking.  Don't give a spherical geometry answer in a course in Euclidean geometry unless you are willing to do a great deal of possibly unsuccessful explaining
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #110 on: June 29, 2021, 02:19:22 pm »
I did some looking for 'g' subscripts ( although my Android smart phone makes subscripts difficult text to enter).
  Looking at some May 13 posts, there are subscripts; (moon), (sun), etc. usually by
   antiprotonboy and also Brumby, so that's helpful.
Plus the process suggested kind of relating back to Earth, as a benchmark.
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #111 on: June 29, 2021, 02:20:10 pm »
I agree with both these last two posts.

I feel some contributors began reading halfway through and felt that they needed to explain the difference between 'g' and 'G'.

To say 'g' is a constant or not I feel can be open for discussion, however, it is based on where you are on Earth, therefore I feel it's more of a calculation and not a constant.

Pi, as an example, doesn't need to be calculated, in fact, it's a button on many calculators. Therefore, it's obviously a constant, but with some long term polishing.
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #112 on: June 29, 2021, 03:43:09 pm »
Ignoring for a while the controversy about little-g being a legitimate constant, I have some observations about weight and mass, and an historical question at the end of this post.
My first formal course in physics, at age 16, was the first time I became aware of the important difference between mass and weight.  Prior to that, everything of interest to me was in terms of weight:  my own weight, the net weight on a box of cereal, etc.  In the US, this also introduced me to the problem of the word "pound", which in physics class was always a unit of force or weight, and the ugly word "slug" for the unit of mass in the same system, which never appeared outside of physics textbooks.  Since the space race was underway, we were all interested in weightlessness, how much one would weigh on the Moon, and other topics in gravitation away from the Earth.
Now to earlier history and laws:  The United States Constitution, Article I, section 8 gives Congress the power to “fix the standard of weights and measurement", where the framers saw the need for uniform measurements across the several States of the Union.  At about the same time, the founders of the Metric System had a similar need to unify French measurements across the new Republic.  Their work lead to the "BIPM", which I believe translates to "International Bureau of Weights and Measures".  Note that in both cases, the language (and further statutes in the US and elsewhere) refers to "weight", rather than "mass".  As one should not go to a physicist for legal advice, one might not trust a lawyer for scientific clarity.
In Principia, Newton defined mass as the product of volume and density.  (Translated from the Latin: "The  quantity  of  matter  is  the  measure  of  the  same,  arising  from  its  density and bulk conjointly".)   The original definition of the gram was 1 cm3 of water (at the temperature of its maximum density), although that was later changed to the prototype kilogram. 
My question:  historically, was this original definition of the gram considered "weight" or "mass"?  Logically, it would be mass, since the corresponding unit of force was already the "dyne".
« Last Edit: June 29, 2021, 04:13:31 pm by TimFox »
 

Online CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #113 on: June 29, 2021, 05:09:50 pm »
 I can't over emphasize the need to know what problem you are working on.  If you are designing a spring scale for the bath room, or calculating how high your model rocket will fly or most of the other problems you will encounter the variation of g over the Earth's surface doesn't matter.  Other parts of the problem will have unknowns greater than the differences.  You terminal velocity when you jump out of an airplane and how fast you reach it will be far more dependent on air density than on g for example.

If you are doing something where the variatíon of g with location matters I hope that your general understanding of the problem is such that you aren't concerned with the existence of a standardized version of the answer.

In much the same way several governing bodies define pi as a rational number (like 3.14 or 3.14159).  This isn't an attack on mathematics, but a way to avoid legal arguments in trade and commerce how many decimal points to use when computing circumferences and areas

 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #114 on: June 29, 2021, 05:49:34 pm »
"Standardized version of the answer":  Is there a legal definition of g?

The classic textbook "Physics", D Halliday and R Resnick, John Wiley & Sons 1966 (one of a series of revisions of this calculus-based freshman physics textbook)  in section 5-6 on systems of units (pp 90 ff), discusses the problem of mass, weight, force, and gravity.   They state that "in the British engineering system....Legally, the pound is a unit of mass but in engineering practice the pound is treated as a unit of force or weight....The pound-force is the force that gives a standard pound an acceleration equal to the standard acceleration of gravity, 32.1740 ft/sec2."  The book later notes that "the acceleration of gravity varies with distance from the center of the earth, and this 'standard acceleration' is, therefore, the value at a particular distance from the center of the earth.  Any point at sea level and 45 deg N latitude is a good approximation".  In this section, the term "g" is not explicitly mentioned.

Moving forward to section 5-8 (pp 93 ff), discussing the vectors W and g, with the scalar mass m, we read "The quantitative relation between weight and mass is given by W = mg.  Because g varies from point to point on the earth, W, the weight of a body of mass m, is different in different localities."  In later examples, the authors are careful to state "at a point where g = 32.0 ft/sec2" or "in a locality where g = 9.80 m/sec2...in a locality where g = 9.78 m/sec2" for quantitative discussions of a particular weight and mass.
This is the standard discussion from the point of view of physics.  This may not matter to backyard rocketry, but is obviously important to interplanetary rocketry.

The point of what many here think of as my nitpicking is that in the traditional British units, where objects are measured by weight and that is converted to mass (pound-force vs. pound-mass), a legal definition of the acceleration of gravity is required, but modern physicists do not call that "g", since the actual value of g varies from place to place.  In metric countries that use SI units, this problem does not occur, since material is sold by mass (in kg), not weight (in N).

Again, I emphasize that the importance of g is that its value applies to all objects in the region of interest, as opposed to older theories where a heavy object drops faster than a lighter object.
« Last Edit: June 29, 2021, 06:22:23 pm by TimFox »
 

Online CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #115 on: June 29, 2021, 06:41:10 pm »
From Wikipedia which has the references. 

I apologize I did forget to subscript the value properly.  The article mentions that this is common

The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2 (about 32.17405 ft/s2). This value was established by the 3rd CGPM (1901, CR 70) and used to define the standard weight of an object as the product of its mass and this nominal acceleration.[1][2] The acceleration of a body near the surface of the Earth is due to the combined effects of gravity and centrifugal acceleration from the rotation of the Earth (but the latter is small enough to be negligible for most purposes); the total (the apparent gravity) is about 0.5% greater at the poles than at the Equator.[3][4]

Although the symbol ɡ is sometimes used for standard gravity, ɡ (without a suffix) can also mean the local acceleration due to local gravity and centrifugal acceleration, which varies depending on one's position on Earth (see Earth's gravity). The symbol ɡ should not be confused with G, the gravitational constant, or g, the symbol for gram. The ɡ is also used as a unit for any form of acceleration, with the value defined as above; see g-force.

The value of ɡ0 defined above is a nominal midrange value on Earth, originally based on the acceleration of a body in free fall at sea level at a geodetic latitude of 45°. Although the actual acceleration of free fall on Earth varies according to location, the above standard figure is always used for metrological purposes. In particular, it gives the conversion factor between newton and kilogram-force, two units of force.


 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #116 on: June 29, 2021, 06:46:40 pm »
The information in this Wikipedia article is consistent with that I quoted from Halliday and Resnick.
Specifically, adding a subscript to g indicates that is measured in a specified location, while I insist that "g" is a general term for the parameter in different locations.
Luckily, we need not worry about this in metrology when we use SI units.
« Last Edit: June 29, 2021, 07:41:43 pm by TimFox »
 

Online CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #117 on: June 29, 2021, 06:47:31 pm »
I will also give the example describing aircraft and spacecraft accelerations in g's.  In this usage the g is always a standard gravity.  And this usage is rarely, if ever, used where precision is required.  But the g is a relatable unit, requiring less mental gymnastics than saying that most pilots lose consciousness somewhere in the neighborhood of 59 to 69 meters per second squared. 
 

Online CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #118 on: June 29, 2021, 07:04:38 pm »
The information in this Wikipedia article is consistent with that I quoted from Halliday and Resnick.
Luckily, we need not worry about this in metrology when we use SI units.

As I read the documents from BIPM, in metrology you don't use g in any of its connotations in metrology.  Only units traceable to standard distance and time.  The only traceability chains authorized are direct tracing to distance and time, comparison to a reference gravitometer which has used the first method, or comparison to an acceleration value at a specific location measured by a gravitometer that has been calibrated by one of the first two methods. 

 The use of g (in any of its forms) is a convenience for practitioners in any number of fields.

Again, horses for courses. 
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #119 on: June 29, 2021, 07:05:45 pm »
In that case, as you properly stated it, "g" is not italic, since it is a unit of measurement, while g is a parameter, equal to the magnitude of the vector g.
When subjected to an acceleration of 10 g along the vertical direction, the victim will weigh 11 times his normal weight.
In college, I did measure g (in southern Minnesota), using a Kater pendulum.  In the ancient textbooks, before electronic timers, the period was measured by placing the Kater pendulum in front of a grandfather clock's pendulum and observing the beat frequency.  In 1969, I used an -hp- 520 vacuum-tube counter, with Nixie tube readout, that was roughly 19 inches cubed, and a simple photocell.
« Last Edit: June 29, 2021, 07:11:31 pm by TimFox »
 

Online Nominal Animal

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Re: Physics Question - ma = mg
« Reply #120 on: June 29, 2021, 08:17:25 pm »
I forget to use the correct (and useful) subscripts too, sorry.

On this forum, you can write say abp, using [sub]..[/sub] and [sub]..[/sub]; these will show the same way in preview as they will in final posts.  abp is written as a[sub]b[/sub][sup]b[/sup] in the editor.   To get the teletype look, I use [tt]..[/tt].  To stop a bracketed term to be interpreted as markup, you can insert a [tt][/tt] or a [i][/i] (which literally means nothing in teletype, or nothing in italics, respectively) in a suitable place (after each [ the forum software considers the beginning of markup, changing each [foo into [[tt][/tt]foo), to break up a markup-looking expression without visual side effects.  It's like adding zero-width spaces: the sequence the server sees is different, but the end result stays the same visually.

This forum also has MathJax support.  For inline MathJax, for example \$E = m c^2\$, you bracket it with \$ on both sides.  (That snippet being written as \$ E = m c^2 \$ in the editor.)  For block expressions, use $$, for example $$ E = m c^2 \tag{1}\label{NA1} $$, to get $$ E = m c^2 \tag{1}\label{NA1} $$where the added stuff lets you create links to it using \$\eqref{NA1}\$ (rendering to \$\eqref{NA1}\$).  The equation label namespace is usually the entire HTML page, so I recommend using labels with a prefix unlikely to be used in other posts.  That way the visible tags can be the same, but the labels keep them separated.

Unfortunately, the forum code lacks the trigger to apply the MathJax translation on preview, so what you see is the bracketed MathJax expressions instead of the rendered results.  At minimum, it'd only need a simple JavaScript call to a library function tacked on to the AJAX completion of the content preview load; for normal posts and thread view, a related call (that does not render source markup, but the intermediate form the forum software munges the source markup into) is automatically triggered by page load completion.  (Implemented this way, it would not exactly match the way the MathJax module in the forum software does it, but the visual result would be the same for the vast majority of users.  The difference would be that preview would lack the MathML translation step done on the server end.  For full support, the MathJax server-side munging would need to be added to the preview server-side processing, plus the rendering trigger added to the AJAX completion.  Here, MathJax is so rare I guess Dave and cohorts don't see it worth the effort.  I'm not familiar enough with SMF innards to have an opinion, but I do want to point out that any such modification is at risk for adding new bugs, so it isn't just laziness; there is always a risk in modifying forum software.)
« Last Edit: June 29, 2021, 09:58:20 pm by Nominal Animal »
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #121 on: June 30, 2021, 01:45:25 am »
These last few posts about how to write 'g' are interesting.

First it shows that smart people need to be precise about how they write, speak, and dislike others who are wrong. It also connects with "discussions" I have with others.

I've noticed keeping my mouth shut is very difficult when I hear in correct statements, or feel the need to critique others.

No matter what value we call 'g', ma can still equal mg. If I accelerated my car from 0 to 9.8m in one-second, then the Force is the mass times 9.8m/s^2 thus making ma = mg.
 

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Re: Physics Question - ma = mg
« Reply #122 on: June 30, 2021, 06:37:14 pm »
On the magnitude and intuition of forces.

Consider Prince Rupert's drops. Toughened teardrop-shaped glass beads, created by dropping molten glass into water.

Because the glass shrinks as it cools, energy gets stored in internal stresses – internal forces pushing each atom of the glass (mostly silicon and oxygen; silica being SiO2 and glass is silica plus impurities) – are immense.  On the bulbous end, the pressure due to this is on the order of hundreds of megapascals; 700 MPa according to a 2016 paper published in Applied Physics Letters.  Again, that force is not exerted by anything external.  There is potential energy stored in the structure, yes, but no dynamic forces involved; only the same static force that say the ground exerts on the soles of your foot when you're standing put and not sinking into the ground, just pushing the atoms in the drop against each other and nothing external.

That is, if you put a Prince Rupert's drop in a padded box, it won't slowly release its structural tension, no more than standing on a granite floor makes it slowly mellow into marshmallow.  Nor will it explode if you put it in vacuum.  It'll just stay as it is, barring stuff like annealing due to temperature changes, structural changes due to ion or cosmic ray bombardment, and so on.

How much is 700 MPa?  Well, 0.1 MPa = 100 kPa = 100,000 Pa = 1 bar by definition, and 1 bar is just slightly under the standard pressure of air at sea level (1.013 bar); or just about exactly the air pressure at 111m elevation at a 15°C temperature.  So, we're talking about seven thousand atmospheres of pressure, give or take.

Such a glass bead certainly does not look like it has anything of the sort locked in its structure in static forces now, does it?  That's because the internal forces, the internal stresses, are perfectly balanced.  They have to be, or the bead would change shape.  The reason cutting the tail makes the drop explode, is that the fracture thus generated (which propagates in the glass bead at five or six times the speed of sound in air, or 5200 - 6800 km/h), unbalances those forces; and no longer in balance the atoms start moving –– in the human scale, in explosive disintegration.

Glass, or even its easier version, silica or silicon dioxide, SiO2, is not hard to simulate on the atom level, but being amorphic, the structure is difficult to get "right" (the way it is in real-world glass or silica).  I don't know if anyone has modeled Prince Rupert's Drops, but their macroscopic size makes it a bit daunting; you'd need serious (distributed) computing power to do it properly.  Smaller simulations can obviously show details (say, a model of a small region of the surface, extending towards the center of the drop), and those have surely been done.

Anyway, on-the-envelope rough estimates (say, 160pm interatomic distance) means that force per atom based on the pressure is on the order of dozen or two picoNewtons (10-12 N).  Doesn't sound much, until you realize a silicon atom only weighs 28 u ≃ 28 × 1.6605 × 10-27 kg ≃ 4.6×10-26 kg.  If that force was applied to a single atom without any opposing forces, then the instant acceleration of that atom would be some forty million million standard gravities, or about 4×1014 m/s2.

"Stupendous" comes to mind, even if there is a typo of a few orders of magnitude in there; but it "feels about right" to me, so I won't re-check – it's just the crudest possible napkin math.  (Oh, and 1 standard gravity ≝ 9.80 m/s².  It's used in some science fiction to describe spaceship accelerations (usually high, as in hundreds of standard gravities of acceleration, since they need inertial dampeners to stop humans from pancaking if you go beyond single-digit standard gravity accelerations, and such high accelerations to get any place or be able to match velocities in any time frame dramatic enough for us humans), and I like how it sounds.)

The amount of energy stored as potential energy in the structure of matter –– and note that this is just the static, structural forces that don't do work since matter generally stays the same; we're not talking about the utterly ridiculous amounts of energy involved in the matter itself due to \$E = m c^2\$ –– and the static forces in it, are really completely outside our intuitive scales.
« Last Edit: June 30, 2021, 06:44:56 pm by Nominal Animal »
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #123 on: June 30, 2021, 07:21:27 pm »
A brief comment about the mass-energy E=mc2 inherent in ordinary matter:
In nuclear reactions and similar problems, only a small part of this energy is available to the consumer, basically due to the small difference between neutron and proton masses, since "baryon number" is conserved.  For nuclei, the protons and neutrons are baryons, so there can be transformations between them.  The proton is the lightest baryon and things stop there, pending discovery of very long decay times that have been theorized but never observed for proton decay.
However, I have a vague recollection from an interesting lecture on black holes I attended in the mid-70's, that if one were in a safe orbit around a black hole, outside the event horizon, and very carefully (adiabatically) and slowly lowered a mass into the black hole on a very strong fishing line, the total energy transferred to the reel would, in fact, be mc2  .
 
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Online Nominal Animal

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Re: Physics Question - ma = mg
« Reply #124 on: July 01, 2021, 12:53:44 am »
The one field where we do already get \$E = m c^2\$ out, is matter-antimatter reactions; specifically, electron-positron annihilation, as used in e.g. positron emission tomography and positron annihilation spectroscopy.

The positrons are generated by nuclear fission (in a radioactive material, called tracer or radiotracer in PET and injected into the patient at very small quantities, and just placed next to the material or structure to be analyzed for PAS).

It just happens that when a positron and an electron annihilate each other at low energies (meaning have low velocities if measured in the frame where their linear momentums cancel out), the end result is two gamma photons with very easily identified energies, and we can use those gamma photons for useful stuff.

At higher energies, by controlling the amount of kinetic energy, you can generate interesting exotic particles like mesons and W and Z bosons, which all tend to decay very quickly (as in within 10-18s or sooner, much much faster than we for example do time steps in atomic simulations) into leptons and hadrons and neutrinos and such.  Interesting stuff for particle physicists, but it's those gamma photons that are very useful in practical applications.

The CERN Antiproton Decelerator, among other thinds, produces antihydrogen: atoms consisting of an antiproton and a positron.  The numbers are small; in 2011, the ALPHA experiment captured 309 antihydrogen atoms for over fifteen minutes (1000 seconds).

In certain Science Fiction settings, antihydrogen is stored as a fuel, and generated in huge accelerator rings, sometimes surrounding planets or stars.  It is likely unfeasible (and something like just sieving positrons and antiprotons from highly energetic environments would work better), but not out of the realm of possibility; not realistic, but close enough to be tantalizing.
 


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