Length Problem

Coordinator
Mar 10, 2008 at 6:04 AM
Length
Coordinator
Mar 10, 2008 at 6:16 AM
My concern about the library is to predict the quantity result from multiply or divide process

to make this Available I store the Dimension as a Key and the Quantity Object in a hash table when the library first start

however the problem of Torque and Work is that the both have the same Dimension which is

F * L Force * Length

When investigating in that problem I come up with the idea to Express L in some more types so its like an internal meaning for the length

if you review the Angle as Quantity

you will find that the angle is the division of two lengths first is the Arc length, second is the radius length

so Basically I thought if L are NL, AL, RL (Normal Length, Arc Length, Radius Length)

Angle will be AL=1 RL=-1 and its L = AL+RL = 0 ==> this is correct as the dimension will be zero or dimensionless
and also the quantity could be saved as a key for the object so I can retreive the object again based on the dimensionless key

when seeing Torque the Toruqe will be F * RL

and it is known that Torque * Angle producing Work which is F =1 RL=1 + AL =1 RL=-1

the produced Dimension is F=1 AL =1 which is expressing the work dimension correctly


Coordinator
Mar 10, 2008 at 7:10 AM
I revised my assumptions

so Length could be two variations Normal Length, and Radius Length NL, RL

I didn't find the need of Arc Length { until now }

so when I made the Angle NL / RL all prediction of quantities were successfull (untill now)

Toruqe * Angle produced energy correctly.
Coordinator
Sep 14, 2008 at 11:41 PM
How are you :)

I just wanted to explain what I made about Angle and Torque in more details

if you divide the Length into NL, and RL :  (Normal Length, Radius Length)   { I want revesion here is it Radius or Radial ?? }

Then you can express  Angle dimensions  as   NL= 1, RL= -1  and L = NL + RL = 0   {zero which mean it is dimensionless}

So
go ahead and define Toruqe

Torque     dimensions      are   Force * Length   (but wait a second :)  the length here is not the Normal Length  it is a Radius Length

remember Work ???     Work was also Force * Length   and that was the main trigger to me about differentiating between length types

so if you redefine Torque on the basis of Force * Radius Length   Then  your   Dimensions on the base of force   are    F = 1,  L = RL = 1


so what about Energy :)


Torque * Angle = Energy    (can't beleive ? :)

(F = 1, L = RL = 1)  + ( L = (NL = 1), (RL -= 1) = 0 )

summing all together the rest will be

F = 1, L = NL + RL - RL  

F = 1, L = NL     ==> which is the Energy  :)   

and by this approach I ended to separate the quantities meanings between Work  and Torque

the question now is there another situations that need reddfine in meaning ?

is my Assumption safe to be considered ?


all this are opened for discussions


and thank you all.