Download 2 - METU | Department of Mechanical Engineering

MECHANICAL
ENGINEERING SYSTEMS
LABORATORY
Group 02
Asst. Prof. Dr. E. İlhan KONUKSEVEN
UNCERTAINTY ANALYSIS
ERROR
IT IS THE DIFFERENCE
BETWEEN THE MEASURED AND
TRUE VALUE
THE TRUE VALUE MUST
BE KNOWN IN ORDER
TO CALCULATE THE
ERROR
Measurement
Uncertainty
Error
Measured Value
True Value
UNCERTAINTY ANALYSIS
UNCERTAINTY
SINCE THE TRUE VALUE IS UNKNOWN
UNCERTAINTY IS THE ESTIMATED ERROR
IT IS WHAT WE THINK THE ERROR IS
UNCERTAINTY ANALYSIS
PRECISION ERROR (Repeatability)
IT IS PRESENT WHEN SUCCESSIVE
MEASUREMENTS OF AN UNCHANGED QUANTITY
YIELDS NUMERICALLY DIFFERENT VALUES
In robotics,
Repeatability: ability of the
robot hand/tool to reach the
same taught position.
UNCERTAINTY ANALYSIS
ACCURACY ERROR
IT IS PRESENT WHEN THE NUMERICAL AVERAGE
OF SUCCESSIVE READINGS DEVIATES FROM THE
KNOWN CORRECT READING
In robotics,
Accuracy: The ability of the
robot hand/tool to precisely
reach a programmed position.
UNCERTAINTY ANALYSIS
ACCURACY ERRORS MAY BE CORRECTED
(FOR EXAMPLE BY CALIBRATION)
PRECISION ERRORS HAVE TO BE
CALCULATED IN A STATISTICAL MANNER
UNCERTAINTY ANALYSIS
WHEN A MEASURED QUANTITY
(X)
IS UNCERTAIN THEN THE RESULT IS
PRESENTED IN THE FOLLOWING
MANNER :

X  X 
 X 
UNCERTAINTY ANALYSIS
IF A MEASUREMENT OF A PHYSICAL
QUANTITY IS EXPRESSED WITHOUT ANY
UNCERTAINTY
THE UNCERTAINTY IS DEDUCED FROM THE
SIGNIFICANT FIGURES
UNCERTAINTY ANALYSIS
SIGNIFICANT FIGURES
THE NUMBER OF FIGURES USED IN
EXPRESSING THE RESULTS OF A
MEASUREMENT IS AN INDICATION OF THE
ACCURACY OF THAT MEASUREMENT
EXAMPLE
IF THE MASS OF AN OBJECT IS SPECIFIED AS
12 Kg
m = 12 kg
THEN THE TRUE VALUE IS ESTIMATED TO
BE CLOSER TO 12 kg
THAN TO 11.5 kg OR 12.5 kg
11.5 m < 12.5
IF THE SAME MASS WAS MEASURED BY A MORE
ACCURATE SCALE AS :
m = 12.0 kg
THEN ITS TRUE VALUE IS ESTIMATED TO BE
CLOSER TO 12.0 kg
THAN TO 11.95 kg OR 12.05 kg
11.95 m < 12.05
Measurements reported as 1 decimeter, 1.0 decimeters, and 1.00 decimeters
may seem the same, but their accuracies are very different.
1 decimeter implies that it could actually range from a half (0.5) decimeter
to just under 1.5 decimeters. It's been rounded to the nearest whole
decimeter.
The ".0" in 1.0 decimeter implies accuracy to the nearest tenth. The actual
length could range from 0.95 decimeters (which got rounded up to 1.0) to
just under 1.05 decimeters (which got rounded down to 1.0)
The ".00" in 1.00 decimeter implies accuracy to the nearest hundredth. The
actual length could range from 0.995 decimeters (which got rounded up to
1.00) to just under 1.005 decimeters (which got rounded down to 1.0)
IN
m = 12 kg
THERE ARE 2 SIGNIFICANT FIGURES
IN
m = 12.0 kg
THERE ARE 3 SIGNIFICANT FIGURES
FOR WHOLE NUMBERS THE
SIGNIFICANT FIGURES ARE
CONSIDERED TO BE INFINITE
FOR EXAMPLE IN THE FORMULA
1
2
E  mv
2
THE 2 IN THE DENOMINATOR IS CONSIDERED
TO BE EXACT (==> 2.0000000…..)
FOR
1.TRAILING ZEROES
E.g. 600 OR 100000 OR 120
2.LEADING ZEROES
E.g. 0.06 OR 0.0001 OR 0.12
TOTAL NUMBER OF DIGITS MAY NOT ALWAYS
CORRESPOND TO NUMBER OF SIGNIFICANT FIGURES
IN ORDER TO CLEARLY EXPRESS THE
NUMBER OF SIGNIFICANT FIGURES USE
SCIENTIFIC NOTATION
FOR EXAMPLE
INSTEAD OF X = 2300
X = 2.3 x 103 ( 2 SIGNIFICANT FIGURES)
X = 2.30 x 103 ( 3 SIGNIFICANT FIGURES)
X = 2.300 x 103 ( 4 SIGNIFICANT FIGURES)
Leading zeros with decimal fractions; e.g.,
x = 0.032 mm (total number of digits is 4) has only 2
significant figures.
IF ONLY SIGNIFICANT FIGURES
ARE USED TO EXPRESS
THE MEASURE OF A QUANTITY
THE UNCERTAINTY IS IN THE
LEAST SIGNIFICANT DIGIT
IF THE UNCERTAINTY IN THE MEASURE OF A
QUANTITY IS OTHER THAN THE LEAST SIGNIFICANT
DIGIT THEN IT IS EXPLICITLY EXPRESSED AS :
*
ABSOLUTE UNCERTAINTY
e. g .
*
RELATIVE
x  125  5
UNCERTAINTY
e. g.
x  125  4 %
COMPUTATIONS OF NUMBERS HAVING
UNEQUAL NUMBER OF SIGNIFICANT FIGURES
ADDITION AND SUBTRACTION
RESULT IS EXPRESSED WITH AN
ACCURACY EQUAL TO THE
ACCURACY OF THE
LEAST ACCURATE NUMBER
COMPUTATIONS OF NUMBERS HAVING
UNEQUAL NUMBER OF SIGNIFICANT FIGURES
ADDITION AND SUBTRACTION
for example
p1 = 18.7 kPa (3 significant figures but less accurate)
p2 = 0.121 kPa (3 significant figures but more accurate)
give
p1+p2 = 18.821 kPa = 18.8 kPa
(1 significant fractional digit)
(3 significant figures )
COMPUTATIONS OF NUMBERS HAVING
UNEQUAL NUMBER OF SIGNIFICANT FIGURES
MULTIPLICATION AND DIVISION
RESULT IS EXPRESSED WITH AN
ACCURACY
EQUAL TO OR LESS THAN
THE ACCURACY OF THE
LEAST ACCURATE NUMBER.
COMPUTATIONS OF NUMBERS HAVING
UNEQUAL NUMBER OF SIGNIFICANT FIGURES
MULTIPLICATION AND DIVISION
for example
F = 172.8 N
v = 0.383 m/s
(4 significant figures & less accurate)
(3 significant figures & more accurate)
give
P = F.v = 66.1824 W = 66.2 W
(3 significant figures)
Propagation of Uncertainty
COMBINATION OF UNCERTAINTIES
AS WELL AS ADDITION, SUBTRACTION,
MULTIPLICATION AND DIVISION THERE
WILL ALSO BE CASES WHERE POWERS OF
VARIABLES WILL BE COMBINED TO OBTAIN
A RESULT.
COMBINATION OF UNCERTAINTIES
THE UNCERTAINTY OF THE RESULT
WILL DEPEND :
*
ON THE UNCERTAINTITIES OF
THE INDIVIDUAL MEASURED
QUANTITIES
*
ON HOW THESE QUANTITIES
ARE COMBINED
COMBINATION OF UNCERTAINTIES
IN GENERAL IF RESULT Q IS A FUNCTION OF MORE
THAN ONE VARIABLE xi
THEN THE EXPECTED VALUE Q WILL BE
CALCULATED THROUGH THE EXPECTED VALUES
OF THE AFFECTING VARIABLES xi
AND WILL HAVE AN OVERALL UNCERTAINTY
Qx 1 , x 2 , . . .  Q 
_
 Q 
COMBINATION OF UNCERTAINTIES
Q x 1 , x 2 , . . .   Q 
_
 Q 
Q = Q(x1+x1, x2+x2, … , xn+xn) - Q(x1, x2, … , xn)
Q
WILL BE DEPENDENT ON  x
AND ON HOW MUCH EACH x i
AFFECTS Q
Q
i.e.
x i
i
The Taylor series expansion of the first term on the right
hand side gives:
Q
Q
Q
Q 
x 1 
 x 2  ... 
x n
x 1
x 2
x n
THE VARIABLES ( xi ) ARE ASSUMED TO BE INDEPENDENT
Maximum Uncertainty:
It is the “ultimate upper bound of the total error” and
can be found as:
Q
(Q) max 
x
i
i 1 x
n
i
Expected Uncertainty:
BY
ADDING
THE
SQUARES
OF
RELATIVE
UNCERTAINTY EFFECT OF EACH VARIABLE AND
THEN TAKING THE SQUARE ROOT OF THE SUM WE
ARE
CONSIDERING
THE
EFFECT
OF
VARIABLE IRRESPECTIVE OF ITS TREND
EACH
  Q


 Q

Q   
x 1   
 x 2   . . .

 x 2

  x 1

2
2
(Q) exp 
Q



x
  x i 
i 1 i

n
1
2
2
Note that always
(Q ) exp  (Q ) max
  Q


 Q

 x 2   . . .
Q   
x 1   

 x 2

  x 1

2
2
1
Q
WE MAY CONSIDER
x i
AS WEIGHING FACTORS THAT
DETERMINE THE RELATIVE EFFECTS
OF x i ON Q
2
Example:
x1 = 7 ± 0.040 cm
z
x2 = 5 ± 0.020 cm
x3
x3 = 5 ± 0.030 cm
x1
x2
The nominal value of z is found as
z = f(x1, x2, x2) = [(x1+x2)2+x32]½ = 13 cm
The respective partial derivatives are calculated as:
x  x 2 12
f
 1

x1
z
13
x  x 2 12
f
 1

x 2
z
13
f
x
5
 3 
x 3
z
13
The maximum uncertainty is calculated as:
(z)max=±[(12/13)*0.040+(12/13)*0.020+(5/13)*0.030)]
= ± 0.067 cm
Hence
z = 13 ± 0.067 (max) cm
The expected uncertainty is calculated as:
(z)exp
= ± {[(12/13)*0.040] 2 + [(12/13)*0.020]2
+ [(5/13)*0.030)]2}½
= ± 0.043 cm
Hence
z = 13 ± 0.043 (exp) cm
Example:
The maximum
stress at the root
is given by
6FL

t
bt2
F
L
b
When dealing with product functions, it is easier to
work with the “relative uncertainty”. Therefore, let
the relative error constituents be known as:
F/F=±0.10;
L/L=±0.05;
b/b=±0.05;
t/t = ±0.08
Then, an expression for the relative uncertainty in the root
stress can be obtained by taking the “ln” first and then
“differential” of its equation given above as:
ln  = ln 6 + ln F + ln L - ln b - 2 ln t
(d/) = (dF/F) + (dL/L) - (db/b) - 2 (dt/t)
Then, the maximum relative uncertainty in  is calculated as:
(/)max = ± (0.10+0.05+0.05+2*0.08) = ± 0.36
and the expected relative uncertainty in  is calculated as:
(/)exp = ± [(0.10) 2 + (0.05)2 + (0.05)2 + (2*0.08)2]½ = ± 0.20
Notice the unexpectedly increased uncertainty level in the
root stress as compared to uncertainty levels given for the
measurements in F, L, b, and t.
EXAMPLE
2
V
P  V. I 
R
V1
R
V  10  0.5 volts
R  100  3 
V2
V1
V2
R
2
V
P  V. I 
R
P
10
EXPECTED
VALUE
2
100
 1 watt
IT IS POSSIBLE THAT
V  10  0.5 Volts
R  100  3 
P
10.5
2
 1.14 watt
97
2
9.5
P
 0.876 watt
103
V1
V2
R
2
V
P  V. I 
R
P
10.5
2
 1.14 watt
97
2
9.5
 0.876 watt
V  10  0.5 Volts P 
103
R  100  3 
WOULD GIVE US
P  1.00  0 .14 watt
V1
V2
R
2
V
P  V. I 
R
HOWEVER WE ARE LESS
CERTAIN OF BOTH
*
THE MEASURED VOLTAGE
AND
*
THE MEASURED OR
ACCEPTED RESISTANCE
V  10  0.5 volts
R  100  3 
THEREFORE
FOR
2
V
P
R
P
V
2
V
R
P
V
 2
R
R
2
P
10
v
2
 0.2

100
V
2
P
10
v

2
2   0 .01

R
100
2
  P

 P

R 
P   
V   
 R


  V
2
P 
0.20.5
2
2



   0 .01 3
P  0.1 Watt
THEREFORE
1
P  1 0.1 Watt
2
2

1
2
EXAMPLE
z  15 x 0.5 z
z  x .y
x  15  0.5
y  0.5  0.05
 y  0.5
x
z
 x  15
y
 z

  z
z   .x    .y 
  y
 x

2

2
z  0.5x0.5   15x0.05 
z  0.8
z  7.5  0.8
2



2
1

1
2
2
EXAMPLE
15
z
0.5
x
z
y
x  15  0.5
y  0.5  0.05
z 1
 2
x y
z
x
  2  60
y
y
 z
  z 
z   .x    .y 
  y 
 x
2
2



1
2
z   2x0.5    60x0.05  


z  3.3
2
z  30  3.3
2
1
2