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Work and Torque (Solution)

April 15, 2010

I am really sorry

I didn’t notice that I haven’t posted how did I solve the dilemma of Torque and Work Problem

however the solution was there in my discussion of quantity System

and also on this post of my personal blog

Problem Core: Length

Simply I defined TWO Length Types

Normal Length (NL)

This is the normal length that we refer to it in our daily life

Radial Length (RL)

This is the radius length that have an origin point in center of circle


Work:       Force * Normal Length

Torque:   Force * Radial Length

Angle:      Normal Length / Radial Length

In quantity system I made the L dimension as NL+RL


I made it explicitly a quantity that is NOT dimensionless

and THIS SOLVED ALL my problems of this Problem

I won’t argue much let us test 🙂

Torque * Angle = Work

F*RL * (NL/RL) =  F * NL    <== see what I mean

Torque * Angular Speed = Power

F*RL * (NL/RL*T) = F*NL/T   <== where T is the time.

so you may wonder about Angle and Solid Angle

in SI they are all dimensionless  but in Quantity System YOU CAN’T consider them like this

(By the way I’ve break the checking for these two quantities to be summable with dimensionless numbers – just to keep the fundamentals as it is although I am not convinced and I may remove it in future but damn it I need support from any physics guy)


Solid Angle:  NL^2/RL^2   because its area over area

Frequency =  1/T

Angular Velocity = (NL/RL)/T

do you remember the conversion between RPM to frequency

RPM (Revolution Per Minute) is a Angle / Time

lets test the law

Omega  = 2*pi*frequency

pi: radian value which is  NL/RL

Omega = (NL/RL) * 1/T = (NL/RL)/T   which  angular velocity


PI value

PI is ratio of any circle‘s circumference to its diameter    WHICH MEANS  (NL/RL)

this is the same value as the ratio of a circle’s area to the square of its radius     WHICH MEANS  (NL^2/RL^2)

which corresponds to radian unit and stradian unit for angle and solid angle quantities.

Reynolds Number

I am not holding back ( Reynolds number is a dimensionless number that measure inertial forces to viscous forces)

WHY we always differentiating between Flow in Pipes and Flow on Flat plate

let’s see with normal length in quantity system

Qs> rho = 3[Density]
Density: 3 <kg/m^3>
Qs> v=0.5<m/s>
Speed: 0.5 <m/s>
Qs> l=4<m>
Length: 4 m
Qs> mue = 2<Pa.s>
Viscosity: 2 <Pa.s>
Qs> rho*v*l/mue
DimensionlessQuantity: 3 <kg/m.s^2.Pa>

it shows it is a dimensionless quantity

ok let us force my theory about flow in pipes

the pipe have a diameter  and Reynolds number is calculated by rho*v*DIAMETER/viscosity

so let me add d as a diameter and solve again

Qs> d=0.5<m!>

RadiusLength: 0.5 m

Qs> rho*v*d/mue

DerivedQuantity: 0.375 <kg/m.s^2.Pa>

NOTE: Adding ‘!’ after the length unit will mark the quantity as Radial Length quantity.

ERROR  it shouldn’t be DerivedQuantity at all it should be Dimensionlesss

so there is another term that should be fixed. Do you know which term ???

lets try the velocity  with vr=0.5<m!/s>

Qs> vr=0.5<m!/s>

DerivedQuantity: 0.5 <m/s>

Qs> rho*vr*d/mue

DerivedQuantity: 0.375 <kg/m.s^2.Pa>


ok let us try the density

Qs> rhor = 3<kg/m!^3>

DerivedQuantity: 3 <kg/m^3>

Qs> rhor*v*d/mue

SolidAngle: 0.375 <kg/m.s^2.Pa>


do you remember the above argue about Solid Angle is dimensionless number

can I pretend now that reynolds number for pipes is CORRECT ???

the new density which is kg/m!^3

This is driving me nuts

Pump Affinity Laws

Also  pump equations led to the same SolidAngle Quantity not DimensionLess one


I’ve said what I’ve discovered till now about this problem I hope that may be someday someone explain to me or convince me that Angle is really a Dimensionless number   after all of this and that my assumption about Torque and Work is Wrong.


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