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Formula for Error Calculations

We have already shown how to calculate Absolute error, mean absolute error, relative error and percentage error in previous posts but the real problem arises when we have two quantities and we need to find out the combined errors.

Here are the formula’s to be used

Lets say we have three quantities a, b and x. The absolute errors in these are respectively + – Δa, + – Δb and + – Δx

Case 1 When x= a+b then Δx= +- [Δa+Δb]
Case 2 When x= a-b then Δx= +- [Δa+Δb]
Case 3 When x= axb then Δx/x= +- [Δa/a+Δb/b]
Case 4 When x= a/b then Δx= +- [Δa/a+Δb/b]

Case 5 When x= an bm/ co then
Δx/x = +- [nΔa/a+mΔb/b+pΔc/c]

posted in Dimensional Formulae | 1 Comment

What is Absolute Error, Relative Error and Percentage Error?

Absolute Error Formula
Absolute error is defined as the magnitude of difference between the actual and the individual values of any quantity in question.

Say we measure any given quantity for n number of times and a1, a2 , a3 …..an are the individual values then

Arithmetic mean am = [a1+a2+a3+ …..an]/n

am= [Σi=1i=n ai]/n

Now absolute error formula as per definition =
Δa1= am – a1
Δa2= am – a2
………………….
Δan= am – an

Mean Absolute Error= Δamean= [Σi=1i=n |Δai|]/n
Note: While calculating absolute mean value, we dont consider the +- sign in its value.

Relative Error or fractional error
It is defined as the ration of mean absolute error to the mean value of the measured quantity
δa =mean absolute value/mean value = Δamean/am

Percentage Error
It is the relative error measured in percentage. So
Percentage Error =mean absolute value/mean value X 100= Δamean/amX100

An example showing how to calculate all these errors is solved below
The density of a material during a lab test is 1.29, 1.33, 1.34, 1.35, 1.32, 1.36 1.30 and 1.33

So we have 8 different values here so n=8

Mean value of density u= [1.29+1.33+1.34+1.35+1.32+1.36+1.30+1.33] / 8 = 1.3275 = 1.33 (rounded off)

Now we have to calculate absolute error for each of these 8 values
Δu1 = 1.33 – 1.29 = 0.04
Δu2 = 1.33 – 1.33= 0.00
Δu3 = 1.33 – 1.34= -0.01
Δu4 = 1.33 – 1.35= -0.02
Δu5 = 1.33 – 1.32= 0.01
Δu6 = 1.33 – 1.36= -0.03
Δu7 = 1.33 – 1.30= 0.3
Δu8 = 1.33 – 1.33= 0.00

Now remember we don’t take +- signs in calculating Mean absolute value
So mean absolute value = [0.04+0.00+0.01+0.02+0.01+0.03+0.03+0.00]/8 = 0.0175 = 0.02 (rounded off)

Relative error = +- 0.02/1.33 =+- 0.015 = +- 0.02

Percentage error = +- 0.015*100 = +- 1.5%

posted in Dimensional Formulae | 29 Comments

How to Convert Units from one System To Another

For using this formula we need to first keep in mind that this formula is used only when the quantities are expressed in absolute units. i.e. in terms of Mass Length and Time.

Terms used in this formula
n1 – Numerical Value in system 1
n2 -Numerical Value in system 2
u1 – Unit of Measurement in system 1
u2 – Unit of Measurement in system 2
M1 – Fundamental Unit of mass in system 1
M2– Fundamental Unit of mass in system 2
L1– Fundamental Unit of length in system 1
L2– Fundamental Unit of length in system 2
T1– Fundamental Unit of Time in system 1
T2– Fundamental Unit of Time in system 2
a = Dimensions of Mass
b = Dimensions of Length
c = Dimensions of Time

Formula to convert is
n2= n1u1/ u2 = n1[M1/M2]a[L1/L2]b[T1/T2]c

Example showing how this formula works
How to convert 1 Newton to Dyne

We know the dimensional formula of Force F = [M1 L1 T-2]

So a =1, b=1, c=-2
Now because Newton is in m.k.s system, we have
M1 = 1kg
L1 = 1m
T1 = 1s

M2 = 1g
L2 = 1cm
T2 = 1s

n1= 1 Newton
n2 = we have to find

Using the formula
n2= n1[M1/M2]a[L1/L2]b[T1/T2]c

n2=1 [1kg/1g]1[1m/1cm]1[1s/1s]-2

n2=1[1000g/1g]1[100cm/1cm]1x1
n2=1000*100 = 100000 = 105

Hence 1 Newton = 105dyne

posted in Dimensional Formulae | 3 Comments

What is the Dimensional Formula of Moment of Inertia?

Moment of Inertia is defined as the product of mass and the square of the perpendicular distance to the specified axis of rotation.

Mathematically,

Moment of Inertia is defined as the product of mass and the square of radius of gyration

Moment of Inertia = Mass X (Radius of Gyration)2

Now we know
Dimensional Formula of Mass= (M1L0T0)
Dimensional Formula of Radius of Gyration= (M0L1T0)

(Radius of Gyration)2=M0 L2 T0

Putting these values in above equation we get
Dimensional Formula of Moment of Inertia= M1L2T0
SI unit
of Moment of Inertia is kg m2

posted in Dimensional Formulae | 2 Comments

What is the Dimensional Formula of Radius of Gyration?

Radius of Gyration is denoted by K and has the same dimensional formula as distance.

Its relation out of Mass, Length and Time is only with Length and thus
Radius of Gyration = Distance ( Dimensionally only)

Dimensional Formula of Radius of Gyration= M0L1T0
SI unit of Radius of Gyration is m

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