RET Destroyed: Objects do not fall at the same rate

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #300 on: December 11, 2012, 07:42:13 AM »
OP:  When values in a math equation change, so do the results.  That's essentially what I got out of it anyways.







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dephelis

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #301 on: December 14, 2012, 06:29:18 PM »
One of the core beliefs of RET is that gravity will cause objects to fall at the same rate regardless of weight. (Given that they are subject to the same gravitational acceleration)

But when you use empirical formulas to calculate terminal velocity, this simply is not the case.

Lets take a perfectly spherical bowling ball, and a perfectly spherical water balloon. We will say that both of their diameters are 8.5in, so that they have the same surface area, and thus have the same wind resistance.

Using an equation derived from the Karamanev Method:

CD = (4/3)*(g*D/ut2)*(ps-pf)/pf

where CD is the constant due to the high turbulence of falling: 0.44

You can calculate then terminal velocities of each of the objects.

Lets start with the bowling ball.

0.44 = (4/3) * (32.2ft/s2*0.708ft)/ ut2 * (86 lbm/ft3-0.07647 lbm/ft3)/ 0.07647 lbm/ft3

You get a terminal velocity of 278.61 ft/s.


Now lets do it again with the water balloon.

0.44 = (4/3) * (32.2ft/s2*0.708ft)/ ut2 * (62.30 lbm/ft3-0.07647 lbm/ft3)/ 0.07647 lbm/ft3

The terminal velocity of it is 237.09 ft/s!

The objects are of the same size and wind resistance, and yet the heavier one falls nearly 50ft/s faster than another!

The gravity hypotenuse claims one thing, but when we use empirical evidence, our findings are very different!

I should come back more often, for all those still wondering at the source of the Karamanev 'Method', I might be able to shed some light. It's some years (18) since I last worked with bioreactors while studying for my MSc in Biotechnology, but here we go.

The Dimitre Karamanev Equation (1996) was published in 1997 in Chemical Engineering Communications (Volume 147, Issue 1) in a paper titled "Equations for calculation of the terminal velocity and  drag coefficient of solid spheres and gas bubbles". The terminal velcocity referred to is the terminal settling velocity of bioreactor slurry within the bioreactor environment.

From my notes of the time the variables in the Karamanev equation are:
CD = Drag coefficient
D = Pipe inside diameter
ut (that should be greek letter mu) = Fluid absolute viscosity
ps (greek rho) = Dry solids density
pf = Fluid density

A few terms like g have been thrown in, but they're meaningless as is this equation when applied to the calculating the terminal velocity of a falling object.

Toodles.

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Tausami

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #302 on: December 14, 2012, 07:05:47 PM »
You have an MS in biotech? I'm considering getting the same. Do you think it was worth it?

OP:  When values in a math equation change, so do the results.  That's essentially what I got out of it anyways.

And yet according to RET they shouldn't.

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EnglshGentleman

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #303 on: December 14, 2012, 09:43:23 PM »
Nice try. On his website Dimitre Karamanev himself acknowledges the use of this equation in text books, for this very purpose, and in fact he has created several equations that are in use.

Quote
calculation of the terminal velocity of freefalling solid particles

Look at this pdf: http://ro.uow.edu.au/cgi/viewcontent.cgi?article=2816&context=theses

Then search "Karamanev" and look at the first result.

Oh crap, what is that equation they show? It is identical to the one I give.  ::)

Go ahead and read the pdf some more if you are lost.
« Last Edit: December 14, 2012, 09:50:25 PM by EnglshGentleman »

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #304 on: December 15, 2012, 06:00:49 AM »
Gravity exerts a force on all objects that depends on their mass (F=m*g), now that water balloon has less mass than the bowling ball, so it exerts a weaker force on the air that it is falling through than the bowling ball. The speed that it needs to fall through the air to null out the gravity pull and the air resistance is lower for the object that has less mass. Your calculations are correct because that's just how gravity and terminal velocity works, though your claim that there's something wrong with that formula\gravity is false.

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Tausami

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #305 on: December 15, 2012, 09:21:16 AM »
Gravity exerts a force on all objects that depends on their mass (F=m*g)

Incorrect.

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #306 on: December 15, 2012, 09:26:44 AM »
http://en.wikipedia.org/wiki/Gravitation

In the most common sense, gravitation or gravity is the agent that gives weight to objects with mass and causes them to fall to the ground when dropped.

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #307 on: December 15, 2012, 10:17:58 AM »
You have an MS in biotech? I'm considering getting the same. Do you think it was worth it?

OP:  When values in a math equation change, so do the results.  That's essentially what I got out of it anyways.

And yet according to RET they shouldn't.

RET says no such thing.  Terminal Velocity takes into account mass.  Why wouldn't it?  The OP is talking about two different things and saying one equals the other. 

In the OP scenario the objects would fall at the same rate, but the object with less mass would stop accelerating before the object with more mass.  However the acceleration is absolutely the same for both objects up until that point.

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Tausami

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #308 on: December 15, 2012, 02:28:27 PM »
http://en.wikipedia.org/wiki/Gravitation

In the most common sense, gravitation or gravity is the agent that gives weight to objects with mass and causes them to fall to the ground when dropped.

Incorrect.

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #309 on: December 15, 2012, 03:15:29 PM »
http://en.wikipedia.org/wiki/Gravitation

In the most common sense, gravitation or gravity is the agent that gives weight to objects with mass and causes them to fall to the ground when dropped.

Incorrect.

Incorrect.

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Tausami

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #310 on: December 15, 2012, 03:37:30 PM »
http://en.wikipedia.org/wiki/Gravitation

In the most common sense, gravitation or gravity is the agent that gives weight to objects with mass and causes them to fall to the ground when dropped.

Incorrect.

Incorrect.

So then why are photons attracted to large aggregations of mass in RET?

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Conker

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #311 on: December 15, 2012, 03:45:39 PM »
http://en.wikipedia.org/wiki/Gravitation

In the most common sense, gravitation or gravity is the agent that gives weight to objects with mass and causes them to fall to the ground when dropped.

Incorrect.

Incorrect.

So then why are photons attracted to large aggregations of mass in RET?

Relativity formula related:

E2=(mC2)+(pC2)

Light does not have mass, but has momentum.
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Tausami

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #312 on: December 15, 2012, 03:50:53 PM »
Exactly. Mass is not the only variable here.

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #313 on: December 15, 2012, 04:09:12 PM »
While we're on subject of light having momentum, how does this work when momentum = mass * velocity?  I'm not doubting as we know light is energy and E = hf only works for waves while light is both a particle and wave, it just seems contradictory and I've never heard it thoroughly explained.


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Thork

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #314 on: December 15, 2012, 04:35:17 PM »
While we're on subject of light having momentum, how does this work when momentum = mass * velocity?  I'm not doubting as we know light is energy and E = hf only works for waves while light is both a particle and wave, it just seems contradictory and I've never heard it thoroughly explained.



You will probably enjoy this
#ws" class="bbc_link" target="_blank" rel="noopener noreferrer">How Much Does a Shadow Weigh?

Its globular propaganda and he makes a few mistakes ... eg his speed of sound is actually the speed of sound in air, not the speed of sound in a solid, but meh. You'll lap this globular nonsense up.
« Last Edit: December 15, 2012, 04:43:17 PM by Thork »

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Tausami

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #315 on: December 15, 2012, 04:45:31 PM »
Vsauce! I love that guy almost as much as I do Hank Green.

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Conker

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #316 on: December 15, 2012, 04:52:47 PM »
Vsauce! I love that guy almost as much as I do Hank Green.
Vsauce, Minutephysics, numerphile and Vilehart master race of youtubers. Great to see something besides Pewdie copypastas.
This is not a joke society.
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You shouldn't be allowed to talk on a free discussion forum.

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #317 on: December 15, 2012, 08:11:44 PM »
Interesting video, but it didn't really answer my question.  Can we presume to understand the energy of a single photon without knowing it's frequency?  I believe that is the most simplistic way I can put it.
« Last Edit: December 15, 2012, 08:22:31 PM by digimonkey »

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Tausami

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #318 on: December 15, 2012, 09:52:59 PM »
Interesting video, but it didn't really answer my question.  Can we presume to understand the energy of a single photon without knowing it's frequency?  I believe that is the most simplistic way I can put it.

Well, no. I don't believe we can,, since the two are so closely related. I may be wrong, however. Why do you ask?

Re: RET Destroyed: Objects do not fall at the same rate
« Reply #319 on: December 15, 2012, 10:44:47 PM »
Exactly. Mass is not the only variable here.

Who said it was?

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dephelis

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #320 on: December 17, 2012, 04:29:36 AM »
You have an MS in biotech? I'm considering getting the same. Do you think it was worth it?

On the basis that I now run an IT department while studying for an OU degree in astronomy, probably not!

I had ethical issues with the two companies I initially worked for, that resulted in my disillusionment of the UK Biotech sector.

Nice try. On his website Dimitre Karamanev himself acknowledges the use of this equation in text books, for this very purpose, and in fact he has created several equations that are in use.

Quote
calculation of the terminal velocity of freefalling solid particles

Look at this pdf: http://ro.uow.edu.au/cgi/viewcontent.cgi?article=2816&context=theses

Then search "Karamanev" and look at the first result.

Oh crap, what is that equation they show? It is identical to the one I give.  ::)

Go ahead and read the pdf some more if you are lost.

Interesting paper, as I had no idea that Karamanev's work had been extended out of the bioreactor. I consider myself duly enlightened and chastised.

However, I note that the equation in the paper and the one you have posted are not identical.

yours is:
CD = (4/3)*(g*D/ut2)*(ps-pf)/pf

the paper states:
vt = ((4/3)*(g*d/CD)*(pp-pa)/pa)^0.5

Note that I'm not querying the differing variable names, rather the different form of the two equations. They give quite different results, although the outcome remains the same. Where does the form you used come from, as I would be interested in reading the paper?

Toodles

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EnglshGentleman

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Re: RET Destroyed: Objects do not fall at the same rate
« Reply #321 on: December 18, 2012, 10:54:39 PM »
Interesting paper, as I had no idea that Karamanev's work had been extended out of the bioreactor. I consider myself duly enlightened and chastised.

However, I note that the equation in the paper and the one you have posted are not identical.

yours is:
CD = (4/3)*(g*D/ut2)*(ps-pf)/pf

the paper states:
vt = ((4/3)*(g*d/CD)*(pp-pa)/pa)^0.5

Note that I'm not querying the differing variable names, rather the different form of the two equations. They give quite different results, although the outcome remains the same.

Toodles

His equation has just been solved in terms of drag coefficient instead of velocity.

Take his equation and raise both sides to the 2nd power.

vt = ((4/3)*(g*d/CD)*(pp-pa)/pa)^0.5
 
=

vt2 = ((4/3)*(g*d/CD)*(pp-pa)/pa)

Now we want to isolate the drag coefficent, so multiply both sides by CD

vt2*CD = ((4/3)*(g*d/)*(pp-pa)/pa)

Divide over velocity.
 
CD = (4/3)*(g*D/vt2)*(ps-pf)/pf

It is now in terms of drag coefficient, and the same equation that I had used.



Where does the form you used come from, as I would be interested in reading the paper?

His equations are actually standard in most recent fluid mechanic books. I will see if I can find a version online, but that may be difficult since these are usually books you buy for uni. It would defeat the purpose if they were free online. :P

I found my text book from a few years ago and scanned relevant page instead. It isn't the best image, but it was the best I could get with my scanner and a thick book. If you look at the bottom, it even references the journal you mentioned.

On the second image it should the same equation in the previous paper I had linked. As I said, mine is just in terms of drag coefficient, but you can also consider on the first page when it has (3/4)CD*Re that the Reynolds number is just dissolved into the right side of the equation.

Fluid Mechanics for Chemical Engineers by James O. Wikles
« Last Edit: January 06, 2013, 02:47:13 PM by EnglshGentleman »