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InuyashaX

-Hokage
Kouhai
What physic work do you have? Post it all here, mabey someone might know the problem.
 

daft27

-nya
Retired
Well, I have something that I'd like to look into. The "spectral review" text that you see on series page was actually a "to-do" for the development of a fansub.tv subsystem used to analyze screenshots.

Here is an IEEE document that I just found on the idea. I was pleasantly surprised that someone did some work already on the idea I had.


So, my homework or project is to figure out how to do this with screenshots on FTV.

http://ieeexplore.ieee.org/iel5/9605/30345/01394329.pdf
 

InuyashaX

-Hokage
Kouhai

daft27

-nya
Retired
IEEE= The world's leading professional association for the advancement of technology. Normally you need membership to access their publications, but I've attached the article so you can take a look at it.
 

noob

±ⁿ°°ɓ±
Retired
*Shrug* well while we're at it, I'll just go ahead and say what my problems are, hoping someone would answer it.




 

Bold

-kenja sama
Retired
QUOTE (noob @ Oct 09 2006, 03:53 PM)

It is hard to read this picture and the first one was un-readable!
====================
'''''''''''''''''''''''''''''|
'''''''A-----------------------G
'''''''''''''''''''''''''''''''''''''''''''''''''|
''''''''''''''''''''''H-------------------------B
'''''''''''''''''''''''|
'''''''''''D-----------C

I assume all horizontal strings are of equal length and attached to the centroid (center of mass) of every fish.

The sum of moments on the A-G rod needs to be 0.
The moment generated by A is Ma = Fa * 120 = 3.4*120=408 N*mm.
So the moment at G needs to be Mg = -408 N*mm in order to get Ma + Mg = 0

Just calculate the sum of moments at G to get the weigth of D.
Mg = -408 = Mg - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...

Calculation of sum of moments is easy and gets you out of nearly any static equilibrium problem. Just make to use the right units. For instance, here notice I used mm everywhere. So the moments are in M*mm. Be carefull not to mix meters in there.

Believe it or not this was fun to do and allowed me to put my mind to something else than the maths I am doing right now.
 

noob

±ⁿ°°ɓ±
Retired
QUOTE (Bold @ Oct 09 2006, 02:13 PM)
Just calculate the sum of moments at G to get the weigth of D.
Mg = -408 = Mg - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...

What??? so what you're saying is

-408 -20*5.4 = D*70 - 20*2.0 -20*5.4 ???

which'd give 1.428571...N ??

(units are not included to prevent further mess and star is used instead of multipication sign.)
 

Bold

-kenja sama
Retired
QUOTE (Bold @ Oct 09 2006 @ 02:13 PM)
Just calculate the sum of moments at G to get the weigth of D.
Mg = -408 = Mg - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...
Oups, little typo, I meant
Mg = -408 = Mh - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...

But that is wrong. See my comment under here.


QUOTE (noob @ Oct 09 2006, 04:26 PM) What??? so what you're saying is
-408 -20*5.4 = D*70 - 20*2.0 -20*5.4 ???
Actually, I forgot something. The 60mm rod needs to be accounted for. I calculated the moments about 2 points (G and H). I can't sum them up like that. Scratch what I just said in my above post

If you assume the length of the vertical strings to be negligeable, you can sum all moments about the point where everything is attached to the ceeling.
 

noob

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Retired
QUOTE (Bold @ Oct 09 2006, 02:49 PM) If you assume the length of the vertical strings to be negligeable, you can sum all moments about the point where everything is attached to the ceeling.
That is what I was supposed to do and many suggested. That is, the sum of the moments of right hand side will equal to left hand side. However, I fail to understand this theory.

Are you simply saying:

Ma = Mb + Mc + Md??
 

Bold

-kenja sama
Retired
QUOTE (noob @ Oct 09 2006, 04:57 PM) That is what I was supposed to do and many suggested. That is, the sum of the moments of right hand side will equal to left hand side. However, I fail to understand this theory.

Ma = Mb + Mc + Md??
Watch out for the signs in moments calculation. The sign is usally the source of errors.

If you calculate the moment about the attachement to the ceiling, then all point on the left would be positive and all the ones to the right negative.
In other words:
Ma + Md + Mc = Mb

The tricky part to visualize here is that the fish D will, if it goes down, push the road near B wich will in turn make fish A go down even. In other words, fish D tends to make the mobile rotate counter clockwise.

Try imagining the mobile with hard pieces of wood vertically instead of strings.
 

noob

±ⁿ°°ɓ±
Retired
QUOTE (Bold @ Oct 09 2006, 03:14 PM) Watch out for the signs in moments calculation. The sign is usally the source of errors.

If you calculate the moment about the attachement to the ceiling, then all point on the left would be positive and all the ones to the right negative.
In other words:
Ma + Md + Mc = Mb
So is that it? does Mb really equal to Ma + Md + Mc?

I thought it was more like

Mb = Ma + Md - Mc

Note how Mb and Mc, both are clock wise and Ma and Md are counter-clock wise... Was I on the right track?
 

Bold

-kenja sama
Retired
QUOTE (noob @ Oct 09 2006, 05:18 PM) So is that it? does Mb really equal to Ma + Md + Mc?

I thought it was more like

Mb = Ma + Md - Mc

Note how Mb and Mc, both are clock wise and Ma and Md are counter-clock wise... Was I on the right track?

If you use the numbers provided for the rods, C is about 10mm left of the center of attachement to the ceiling. So C will tend to make the system rotate counter clockwise around the point of attachement.

B is the only one to the right. So it is the only one that will tend to make it turn clockwise. What makes it really weird is that D and C are attached to a point to the right. Try redrawing it with a single stick and all fishes in it

-----------+-------------
________|_________
A....D..C................B

Does it help to see it? Keep in mind that moments an be moved around. The "path" from where the force is applied and the point about which it is calculated is not important.


Note that to make this work in real life, you would ALSO have to make sure the sum moments of each horinzontal bars are 0. But in this case, since you have only one fish to adjust and 3 bars, there is no choice but to calculate for the top bar and hope the others are also stabilized due to a cleaver writer of the problem.
 

noob

±ⁿ°°ɓ±
Retired
That is what I thought I should do too, however according to many, including my instructor, it cannot be assumed that all the other bars have a moment of zero... :/
 

Bold

-kenja sama
Retired
QUOTE (noob @ Oct 10 2006, 01:34 AM) That is what I thought I should do too, however according to many, including my instructor, it cannot be assumed that all the other bars have a moment of zero... :/
You can calculate one equation for each "level". So you will end up with 3 equation but only a single variable. So, basic multi-equation solving, either the same values solves all 3 equations or the problem is not solvable. You have more contrainst (equation) than variable.
 

noob

±ⁿ°°ɓ±
Retired
Ah, I found a shorter way and was able to get the answer. What you are saying is also correct, however with the knowledge I currently have, doing that is nearly impossible. If I remember correctly, the purposes of questions like these are to give you a huge amount of work and expcet you to do them quickly and use different methods.


Ma/30=B+C+D

and since I was given the B, C, and A, I was able to solve for D. Thanks for the help though
 

Bold

-kenja sama
Retired
QUOTE (noob @ Oct 11 2006, 12:19 AM) Ma/30=B+C+D
Yes, you are quite right about that one. I guess my physics are more rusted than I thougth they were!

But you again assume the smaller rods are in equilibrium (that their sum of moments is zero). Nothing wrong with this assumption since this is a text book problem. Not a real world one. (If it was a real world one, you could play with the weigth of the other fishes and with the weigth of the rods to make this work
)
 

noob

±ⁿ°°ɓ±
Retired
No, if the smaller rods were in quailibrium I would not have gone through so much of trouble, trying to find D... lol

In that case it'd be something like

Md = Mc

Dividing Mc by the distance of D would give you Fd.

[ 20 * 2 ] / 70 = Fd = Weight of D


However, using that method, you'll reach an answer of.. 0.5714, no where near the answer using the other method, which would be 6.2... I know, this is quite tricky
 

wittyfox

→√wittyfox=god Since god²=ME
Retired
Any one care to explain to me the diffence of a Raster image and a vector image. I just learned it today in class, but need some refreasing.
 

Kit-Tsukasa

-desu
Retired
QUOTE (wittyfox @ Oct 11 2006, 12:47 AM) Any one care to explain to me the diffence of a Raster image and a vector image. I just learned it today in class, but need some refreasing.
Raster image: usually rectangular grids of pixels making an image look not very smooth when resizing, thus the effects of bluriness in computer images.

Vector image: they are somewhat the opposite of raster images. When creating a vector image, it creates smoother images so that you don't see the rectangular grids of pixels when resizing to any form, size, or shape.

I know it was a very good description, but I hope this helps somehow or clarify things if you didn't understand. By the way, you can try this in photoshop I believe and you can see the difference.
 
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