The use of dimensions or units in physics may be used to
check the homogeneity of physical equations, or obtain a useful empirical
equation from basic measurements, observations.
Example:
Consider a volume V of a liquid flowing per second through a pipe under
steady pressure as shown below:
Fig. 1: The fluid flow
through a uniform pipe
It is reasonable to assume, in the first
instance, that V (volume flow per
unit time) depends upon:
a) The coefficient of viscosity, h, of the liquid
b) The radius, r, of the pipe
c) DP/L the pressure gradient arising from the
flo
Then we may write: V = k hx ry (DP/L)z
Where k denotes a constant, and x, y and z are
indices whose values are to be found.
We know:
V has dimensions L3 T - 1
h has
dimensions M L-1 T -1
r has dimension L
DP has
dimensions M L-1 T - 2
So, DP/L has dimensions: M L-2 T - 2
Then, equating dimensions:
L3 T – 1 = [M L-1 T -1]x
[L]y [M L-2 T - 2]z
We next equate indices for M, L, and T on both
sides:
For M:
0 = x + z
For L: z = -x + y – 2z
For T: -1 = -x – 2z
Next, solve for the indices:
a) x = -z
b) -1 = -(-z)- 2z = -z or z = 1
c) -1 = -x -2z
d) 3 = 1 + y -2 or y = 4
Finally:
V = k h- 1 r 4 (DP/L)1
Or: V =
k DP r 4
/h L
We note here that k cannot be obtained by the
method of dimensions, but a fuller analysis would reveal k = p/8, so:
V = p DP r 4
/8 h L (Poiselle’s formula)
Problems:
1) Use the method of dimensions to obtain a formula for the force
experienced by a sphere of radius r moving at velocity v through a fluid of
viscosity h. (Hint: Set out the force as F = k hx v y r z )
2) Repeat the exercise above to obtain the
period (T) for a pendulum’s swing.
- Errors & Uncertainties in
Measurements
In general, there are two types of errors with
which we will deal in physics:
1) Systematic: an error repeated due to a fault in the instrument of the procedure.
2) Random: an error which arises due to intrusion of some random factor,
or the necessity to estimate and the limits of instrumental sensitivity.
An illustration of a systematic error is
illustrated below for an ammeter. We
observe that in this case, the meter is incorrectly calibrated since the real
zero is displaced from the value read by the ammeter, by and amount 0.2 A. Thus
when the meter reads 1.2 A it is actually 1.0 A.
Fig. 2: Improperly calibrated
ammeter.
In this case, X = 1.0 A represents the “true
value” for the measurement, and the reading (x) is registered as:
x = X + e
Then: +
e =
x – X = 1.2A – 1.0A = +
0.2A
The nature of a random error is illustrated in
Fig. 3.:
Fig. 3: A random error
arising from multiple deviating readings.
In this case, reading of the ammeter yields a
series of slightly different readings at different times, say: x1, x2, x3, and
x4. In this case, x denotes the “true value” and we have:
x1 = x – c1
x2 = x – c2
x3 = x – c3
where c1, c2, and c3 are “correction factors”
of the observed value to the true value, x. In such an instance, we generally
reduce the effects of these random different values by taking the average value
x by:
x = (x1 + x2 + x3 +…..xn)/
n
Which should be closer to the true value x,
than any one of the other observed values.
In general the associated error will be:
e = (x – x )
So the fractional error is: f = e/ x
And the percentage error is: p = (e/x) x 100%
Usually: f < < 1 and p < < 100%
Example:
As illustrated below, a measurement of 2.5 cm
is made using a scale rule graduated in millimeters (mm or 0.1 cm) What would
the possible error (p.e.) be in this case, and how would we write the length of
the object?
Fig. 4.: Example illustrating
possible error of measurement for an object
The possible error in this case is the
equivalent of the limit of accuracy for the scale or 0.1 cm. Then:
p.e. = +
0.1 cm
The length of the object is written: L = 2.5 +
0.1 cm
The percentage possible error (p.p.e.) is:
(+ 0.1 cm/ 2.5 cm) x 100% = + 4%
Example (2):
For the same measurement a vernier caliper may
find:
L = 2.46 + 0.01 cm
Since its accuracy is a factor 10 enhanced
over the scale.
The possible error is then: p.e. = +
0.01 cm
And: the percentage possible error is:
p.p.e. = (+ 0.01 cm/ 2.46 cm) x 100%
= + 0.4%
3. Combining Errors of
Measurements
It is possible and indeed, frequently
necessary, to combine errors. We consider several cases below:
Case (1): The
error sum Q = a + b
Then: The total p.e. in Q = p.e. in a + p.e.
in b
(Also, for absolute errors: DQ = Da + D b)
Let: a = 5.1 + 0.1 cm and b = 3.2 + 0.1 cm
Then Q = (5.1 + 0.1 cm ) + (3.2 +
0.1 cm ) = 8.3 + 0.2 cm
In addition:
Q max = (5.2 + 3.3) cm = 8.5
cm, and
Q min = (5.0 + 3.1) cm = 8.1 cm
Case (2): The
error difference Q = a – b
If Q = a – b then the same rule applies, in
terms of addition of possible errors. Also: Da = D b = +
e)
Case (3) Product
or quotients and fractional errors.
If a quantity x is measured and recorded as x’
units, then let: (x’ – x) denote the error in x or x’ = x + e. But we may also
write:
x’ = x(1 + f)
where f = e/x is the fractional error in x.
Note also that: (x’/ x) = 1 + f.
For a sum or difference: Q = a + b = (a + e1)
+ (b + e2)
And: f = (e1 + e2)/ (a + b)
Also:
f1 = e1/ a and f2 = e2/b
Therefore: f = f1 + f2
Or f = (af1/ a) + bf2/b)
Error in a product, viz, Q = ab
Then: Q = a (1 + f1) x b (1 + f2)
Or: Q = ab(1 + f1 + f2)
Error in a quotient, viz. Q = a/b:
Then: Q = a(1 + f1)/ b(1 + f2)
Q = a/b [ (1 + f1)(1 – f2 + …) = a/b [1 + f1 –
f2]
Problems for the ambitious reader :
1) Two lengths ℓ1 and ℓ2 are measured as (10.0 + 0.1 cm ) and (9.0+ 0.1 cm ). Find the p.e. for each length. Express the
sum (ℓ1 + ℓ2) with errors-uncertainties. Thence or otherwise obtain the
fractional error derived from the sum.
2) The length of a sheet of paper is measured
as: ℓ = 11.0 + 0.1 cm, and its
width as w = 8.5 + 0.1 cm. Find
the Area A = ℓ x w in terms of the associated possible errors. Find the
fractional error associated with the area.
3) A glass thermometer is scaled in tenths of
a degree. If a student reads the lab temperature as 30.5 o C,
express the possible error in the measurement.
