1. DEPARTMENT OF MECHANICAL AND MANUFACTURING ENGINEERING
FACULTY OF ENGINEERING
UNIVERSITY OF RUHUNA
ME 6302 Automatic Control Engineering
ASSIGNMENT NO 01
(DESIGN AND ANALYSIS A MAGNETIC LEVITATION SYSTEM
USING J-MAG DESIGNER AND MATLAB)
Diameter of the ball : 22mm
Selected permanent magnet : N50 (Neomax reversible)
NAME : WEERASINGHA P.K.C.P.
REG NO : EG/2014/2566
DATE : 03/09/2017
2. ABSTRACT
Overcoming the grip of the Earthβs gravity has been a major challenge for years. However, the
work of scientists and engineers who have found many ways to levitate a variety of objects is
being applied in the field of transportation in several countries. Maglev is the means of floating
one magnet over another. According to a theorem attributed to Earnshaw, it is impossible to
achieve static levitation using any combination of fixed magnets and electric charges. There are,
however, ways to levitate by getting around the assumptions of the theorem. Magnetic levitation
employs diamagnetism. In this report consist with magnetic levitation experiment for 350g ball
from ss400 material. From β3π΄ to +3π΄ current with 0.25π΄ step and design for 25 cases. Using
magnetic levitation result from J-Mag Designer, analysis with MATLAB.
3. ii
ACKNOWLEDGEMENT
Under the module ME6302 automatic control engineering, we follow the control system
behavior magnetic lavation is one of this control system. As an assignment completed this report.
It is very pleasure to have the experiences and guidance during completing of assignment task
Dr. Buddika Annasiwaththa, lecturer at Department of Mechanical and Manufacturing
Engineering, Faculty of Engineering Ruhuna. We would like to give our gratitude for leading us
by teaching how to do a project according to the procedure. I have taken efforts in this assignment
with the help of many persons. However, it would not have been possible without the kind
support and help of them. I would like to extend our sincere thanks to Dr. Buddika Annasiwaththa
and all others who helped us.
Thank you,
Weerasingha P.K.C.P.
EG/2014/2566
Faculty of Engineering
University of Ruhuna.
4. iii
TABLE OF CONTENTS
Abstract..........................................................................................................................................i
Acknowledgement........................................................................................................................ii
Table of Contents ....................................................................................................................... iii
List of tables .................................................................................................................................v
List of figures ..............................................................................................................................vi
1 Magnetic levitation system in control engineering ...............................................................1
1.1 Magnetic Levitation Defined .........................................................................................1
1.2 Introduction....................................................................................................................1
1.3 Assignment Task............................................................................................................2
1.4 Manufacturer details for component in magnetic levitation system..............................2
2 Design a simple zero power controlled magnetic levitation system .....................................4
2.1 2D model........................................................................................................................4
2.2 3D model........................................................................................................................5
3 Simulation .............................................................................................................................5
3.1 J-mag Designer ..............................................................................................................5
3.2 Simulation process.........................................................................................................6
3.3 Plotting graph...............................................................................................................10
3.4 The zero-power levitation air gap (ππ) of the SS400 ball ...........................................12
4 Curve fitting tool in MATLAB R2016a..............................................................................18
4.1 Force equation..............................................................................................................18
4.2 Calculating constant and Root Mean Square Error (RMSE).......................................20
4.3 Calculation of theoretical force....................................................................................21
4.4 Difference between calculated force and J-Mag Designer force.................................27
5 Magnetic flux density plot...................................................................................................33
5.1 βπ π¨ and π. π ππ airgap ...........................................................................................33
5.2 +π π¨ and π. π ππ airgap ...........................................................................................34
5.3 π π¨ and π. π ππ airgap ..............................................................................................34
5.4 βπ π¨ and π. π ππ airgap ...........................................................................................35
5. iv
5.5 +π π¨ and π. π ππ airgap ...........................................................................................35
5.6 π π¨ and π. π ππ airgap (zero power levitation air gap (ππ)) ....................................36
6 Simulation details................................................................................................................36
6.1 Mesh.............................................................................................................................36
7 Computer characteristics.....................................................................................................38
8 Linearize the equation .........................................................................................................38
9 Discussion ...........................................................................................................................40
References ..................................................................................................................................42
6. v
LIST OF TABLES
Table 1. AWG copper wire selection ...........................................................................................3
Table 2. Calculate the total resistance and length of copper coil .................................................8
Table 3. Created cases for the design .........................................................................................10
Table 4. Relationship between force and airgap length..............................................................13
Table 5. J-Mag data ....................................................................................................................14
Table 6. Calculated theoritical force...........................................................................................21
Table 7. Time taken to complete each case................................................................................37
7. vi
LIST OF FIGURES
Figure 1. magnetic levitation system model.................................................................................1
Figure 2. N50 1/2"x1" Neodymium Rare Earth Cylindrical Magnet...........................................2
Figure 3. 0.65 mm copper wire.....................................................................................................3
Figure 4. 2D drawing of system ...................................................................................................4
Figure 5. 3D model.......................................................................................................................5
Figure 6. 1st
round of coil .............................................................................................................6
Figure 7. 2nd
round of coil ............................................................................................................7
Figure 8. Electromagnetic circuit .................................................................................................8
Figure 9. After creating mesh of model........................................................................................9
Figure 10. (a) Quantity of mesh given by J-Mag Designer, (b) Quality of mesh of model .......10
Figure 11. The graph of Length Vs. Force acting on Y axis......................................................11
Figure 12. The graph of Length Vs. Force acting on Y axis with reference line as weight of ball
....................................................................................................................................................12
Figure 13. Surface fitting graph for force equation....................................................................19
Figure 14. Surface fitting graph for calculated force..................................................................26
Figure 15. Difference between absolute force and calculated force...........................................32
Figure 16. Flux density in the model β3 π΄ with 1.0 ππ air gap ..............................................33
Figure 17. Flux density in the model +3 π΄ with 1.0 ππ air gap ..............................................34
Figure 18. Flux density in the model 0 π΄ with 1.0 ππ air gap .................................................34
Figure 19. Flux density in the model β3 π΄ with 4.3 ππ air gap ..............................................35
Figure 20. Flux density in the model +3 π΄ with 4.3 ππ air gap ..............................................35
Figure 21. Flux density in the model 0 π΄ with 4.3 ππ air gap .................................................36
Figure 22. Linearization of equation (C)....................................................................................39
Figure 23. cure fitting for J-Mag data with polynomial function...............................................40
8. 1
1 MAGNETIC LEVITATION SYSTEM IN CONTROL ENGINEERING
1.1 Magnetic Levitation Defined
Magnetic levitation is cited on Microsoft Academic Search as follows: βmagnetic levitation is a
technique to suspend an object without any support other than that of a magnetic fieldβ. The
definition goes on to state, βIn a simple configuration, an electromagnet is used to lift up a
ferromagnetic object and the gravity pulls down the objectβ. Stated another way, current passed
through electromagnets creates an attractive magnetic levitation force. Magnetic levitation is also
referred to as magnetic suspension[1].
1.2 Introduction
Magnetic levitation is becoming widely applicable in magnetic bearing, high-speed ground
transportation (bullet trains), vibration isolation, flywheels and levitation melting, household
fixtures and decorations, magnetic levitation in universities, etc. magnetic bearing support radial
and thrust loads in rotating machinery. Also, magnetic suspension generates levitation action is
rectilinear motion devices such as high-speed ground transportation system[1].
The portable magnetic levitation system (MagLev), shown in figure 1, design and analysis using
J-mag Designer and MATLAB software and plot the graph between the current flowing in
copper coil, air gap between ss400 steel ball and the force action on ball when the ball levitates
in air by the magnetic force generated by the
β’ Electromagnet
β’ Permanent magnet
β’ Or both permanent and electromagnet
Figure 1. magnetic levitation system model
9. 2
1.3 Assignment Task
Under the module ME 6302 automatic control engineering, use computer aided design and
analysis tools (Jβmag Designer and MATLAB) to identify the relationship between force, current
and air gap of a simple zero power controlled magnetic levitation system.
1.4 Manufacturer details for component in magnetic levitation system
In this designing process first calculate the radius of the ball and dimension of the permanent
magnet which is available in the market.
1. ss400 ball
β’ Radius of the ss400 ball =
70+150
10
= 22 ππ
2. Magnet;
β’ Radius = 1/2 πππβ
β’ Height = 1 πππβ
Figure 2. N50 1/2"x1" Neodymium Rare Earth Cylindrical Magnet[2]
Custom details
β’ Material : Sintered NdFeB. Grade N50
β’ Gauss Ratings : 14,200 Gauss
10. 3
β’ Pulling Force : 51 lbs.
β’ Pole Orientation : Axially magnetized. Poles on the flat surfaces.
β’ Coating : Ni + Cu + Ni triple layer plated, the best coating available
β’ Tolerance : The tolerances of all the dimensions are +/- 0.002in with coating.
Magnet available in CMS magnets
3. Copper coil
In this design need to flow maximum 5A current to the its coil. But the duration of the current
flowing is very small due to that reason the heat dissipation from the coil is less. Then consider
American Wire Gauge (AWG) to get suitable diameter for the copper coil wire[3].
Table 1. AWG copper wire selection
AWG gauge Conductor
diameter (mm)
Ohms per
Km
Maximum amps
for chassis
wiring
Maximum amps
per power
transmission
22 0.64516 52.9392 7 0.92
23 0.57404 66.7808 4.7 0.729
The 5A between AWG gauge number 22 and 23, therefore used gauge number 22 to the design.
Figure 3. 0.65 mm copper wire[4]
Custom details
β’ Metric Wire Size : 0.65ππ
11. 4
β’ Net Wire Weight : 50 π πππ πΆπππ
β’ Length on spool : 16.5 π
β’ Cross Sectional Area : 0.33187 ππ2
β’ Temperature Rating : 212 Β°πΆ (Class 'H')
β’ Voltage Break Through : 11000 π Minimum Dielectric Strength
β’ Grade : Dual Coat
β’ Base Coating : Polyester Imide
β’ Top Coating : Polyamide Imide
β’ IEC Standards : IEC 60317-13 / DIN 46416-7
Copper coil available in eBay
2 DESIGN A SIMPLE ZERO POWER CONTROLLED MAGNETIC LEVITATION
SYSTEM
2.1 2D model
Based on the above calculation create a 2D sketch for the magnetic levitation system.
Figure 4. 2D drawing of system
12. 5
β The plastic bobbin is design for 3D printing and its vertical axis thickness of 0.50 ππ
β For the bolt, it is design for custom design and it can create from machining by ss400
bar.
β The initial air gap between magnet lower surface and top surface of ss400 ball is
10.0 ππ.
2.2 3D model
Then using 2D sketch created 3D solid model using SolidWorks R2016a
β Take axis of the bolt is Y (then ball levitate through the Y axis).
β Take origin as reference point of model.
β For the levitate propose ss400 ball set to motion region from magnet lower surface
1.0 ππ to 10.0 ππ.
β Create ACIS.sat file format for import to J-mag designer.
Figure 5. 3D model
3 SIMULATION
Basically, used J-mag Design 16.0 to magnetic simulation and create graph between force acting
on the ball and the airgap length.
3.1 J-mag Designer
Special simulation tool used for simulate the
β Magnetic
β Thermal
13. 6
β Structural
β Electric
β Thermal stress
β Transformer
In this deign we used magnetic transient analysis, because it is need to calculate force and air
gap of the model. It can directly import solid works file to the J-mag designer.
3.2 Simulation process
β’ Step 1
After importing the solid model to the J-mag designer, define material for each part of the design;
β’ Ball and bolt : ss400
β’ Bobbin : Plastic
β’ Coil : Copper
β’ Magnet : Neomax reversible N50
for the ball and bolt should be allowed the eddy current because when flux is acting on the metal
surface its create eddy current inside it. But coil surface is coated then itβs no need to allow eddy
current.
β’ Step 2
In the system has electromagnet, for create electromagnet used copper coil. The coil resistance
should be calculated. The coil has 700 turn each one round has 50 turns, then it has 14 rounds.
β’ 1st
round of copper wire
the diameter between 1st
round is sum of the bobbin
outer diameter and two times of radius of the coil, it is
shown figure 6.
Diameter in one coil turn = ππ
= π π₯ 22.65 ππ
Figure 6. 1st
round of coil
14. 7
= 71.1570736 ππ
Bobbin has 50 turns in one round. Then,
Total length of one round = 50 π₯ 71.1570736 ππ
= 3557.85368 ππ
β’ 2nd
round of copper wire
The diameter between 2nd
round is sum of the
bobbin outer diameter and two times of radius
of the coil, it is shown figure 7.
Diameter in one coil turn = ππ
= π π₯ (22 π₯ 1 + 1) π₯ 0.65 ππ
= 59.53318079 ππ
Bobbin has 50 turns in one round. Then,
Total length of one round = 50 π₯ 59.53318079 ππ
= 2976.659039 ππ
According to the figure 7, the total length of each one round can be calculated 2nd
round to 20th
round by taking n = 1, 2, β¦, 19.
The resistance of the unit lent of the copper wire of 22 AWG gauge in metric standard is 52.9392
Ohms per Km.
Then further calculation is shown in table 2 and it gives
Total length of coil = 94.24777961 π
Total resistance of coil = 4.989402054 πΊ
Figure 7. 2nd
round of coil
15. 8
Table 2. Calculate the total resistance and length of copper coil
Number of
round
Diameter Length of one
turn
Number of turn in
each round
Total length of one
round
1 22.65 71.15707360 50 2772.455517
2 23.95 75.24114405 50 2976.659039
3 25.25 79.32521450 50 3180.862562
4 26.55 83.40928495 50 3385.066084
5 27.85 87.49335540 50 3589.269607
6 29.15 91.57742585 50 3793.473129
7 30.45 95.66149630 50 3997.676652
8 31.75 99.74556675 50 4201.880174
9 33.05 103.8296372 50 4406.083697
10 34.35 107.9137077 50 4610.287219
11 35.65 111.9977781 50 4814.490742
12 36.95 116.0818486 50 5018.694264
13 38.25 120.1659190 50 5222.897787
14 39.55 124.2499894 50 5427.101309
Total length(mm) 68392.47207
Total length(m) 68.39247207
Ohms per Km 52.9392
Total resistance (π΄) 3.620642757
The coil should be contained the number of turn and resistance of the coil.
Figure 8. Electromagnetic circuit
β’ Step 3
The initial condition should be applied on the model. There are three basic conditions,
1) FEM coil
2) Translation motion of ball
3) Nodal force action on the ball
16. 9
When applied the FEM coil it should be link to the created circuit in step 2 and also give to the
initial starting point of current flowing through the coil.
When the translation motion applied on the ball is tend to move through the Y axis. The motion
is taken by 0.1 ππ step taking throughout the whole moving distance of ball. As an example, in
this model we take 1ππ to 5 ππ for motion region. If it has 0.1 ππ step total steps of 51 with
first step. In this model applied downward motion and therefore the graph of force on Y Axis
Vs. length (mm) are shown in figure 11.
Nodal fore link to the translation motion.
β’ Step 4
The accuracy of the calculation depends on the mesh of the model, when it has small mesh
element the accuracy of the output is higher. For that model used 1.0 ππ and air element which
contain in control volume to the calculation is 2.0 ππ.
Figure 9. After creating mesh of model
This model has
β 268 754 elements
β 52 676 nodes
β Total element of 321 430
17. 10
(a) (b)
Figure 10. (a) Quantity of mesh given by J-Mag Designer, (b) Quality of mesh of model
β’ Step 5
In this model, used different current value and it is starting β3π΄ to +3π΄ with 0.25π΄ difference
then it has 25 different current value. Each one take as case, then it has 25 all cases. 13th
case
with 0 π΄ current.
Table 3. Created cases for the design
case
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
I(A)
-3.00
-2.75
-2.50
-2.25
-2.00
-1.75
-1.50
-1.25
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.3 Plotting graph
One of main output of the simulation is the graph of force action on ball through Y Axis Vs.
length through Y axis from initial point which has 1.00 ππ away from magnet. First consider
case 13th
, and it should be cut within 3 ππ to 6 ππ region. And after 1st
and 2nd
cases are run
because of the get clear idea about the saturation of the model with the magnetic flux. If it is
saturated, should be change the physical parameter of the model. These parameter is.
β Number of turn of coil
β Magnet diameter
β Diameter of bolt, etc.
18. 11
Figure 11. The graph of Length Vs. Force acting on Y axis
The weight of the ball =
4
3
π₯ π π₯ π3
π₯ π π₯ π
=
4
3
π₯ π π₯ (0.022)3
π₯ 7860 π₯ 9.81
= 3.439126932 π
19. 12
3.4 The zero-power levitation air gap (π π) of the SS400 ball
It can apply reference line to the graph to get length when force is 3.43913 π.
Figure 12. The graph of Length Vs. Force acting on Y axis with reference line as weight of ball
It also has these types of graph for all 25 cases. When ball have 3.43913 π and length of X axis
of reference value for the Y is given by table number 4.
From 1st case to 22nd
case when force is 3.43913 N, curve intersect between 3.00 mm to 6.00mm
region according to the above table. But after 22nd
case intersecting point goes to right hand side
to X axis.
Using these 25 graphs get the reading of airgap length (mm) where 0.1mm each step from 1.00
mm to 6.00 mm and force of each step. These step is previously defined as 51 steps for total
motion region.
20. 13
Table 4. Relationship between force and airgap length
Case number Length of graph (mm) Airgap length (mm)
1 2.86073 3.86073
2 2.82835 3.82835
3 2.80681 3.80681
4 2.79684 3.79684
5 2.79862 3.79862
6 2.81229 3.81229
7 2.83790 3.83790
8 2.87467 3.87467
9 2.92925 3.92925
10 3.00363 4.00363
11 3.07115 4.07115
12 3.14929 4.14929
13 3.24947 4.24947
14 3.36787 4.36787
15 3.49870 4.49870
16 3.65361 4.65361
17 3.82482 4.82482
18 4.00651 5.00651
19 4.20457 5.20457
20 4.43227 5.43227
21 4.67430 5.67430
22 4.93335 5.93335
23 - -
24 - -
25 - -
After the simulation process, it created the 25 graphs. The graph has
β X axis as length (it gives 0 ππ to 5 ππ in positive direction; because initially set the
motion direction as the βπ direction).
β Y axis gives the Force in Y direction
Using that graph get force value in each 0.1 ππ step of each one of graph. Then for every 25
graphs contain this force value and it can create 51 π₯ 25 matrix
β 25 - β3π΄ to +3π΄ with 0.25π΄ time step.
β 51 β 1.0 ππ to 6.0 ππ with 0.1 ππ time step includes with 1.0 ππ.
27. 20
4.2 Calculating constant and Root Mean Square Error (RMSE)
After surface fitting, it gives the details for above equation (B)
General model:
f(x,y) = a*((x + b)^2)/((y + c)^2)
Coefficients (with 95% confidence bounds):
a = 0.3185 (0.2947, 0.3423)
b = 20.9 (20.15, 21.65)
c = 1.149 (1.114, 1.184)
Goodness of fit:
SSE: 2029
R-square: 0.9732
Adjusted R-square: 0.9732
RMSE: 1.263
in this experiment Root Mean Square Error (RMSE) is 1.263, but this value is higher than which
excepted value. Because if the physical model which used in the simulation. If this value can be
lower changing,
β physical dimension of bolt
β turn of the copper wire, etc.
40. 33
5 MAGNETIC FLUX DENSITY PLOT
Considering the all 25 cases it takes -3A to +3π΄ for 0.25π΄ current step with 1.0 ππ which 1st
step in length create the magnetic flux density shown in bellow figures. In this figure is shown
the maximum and minimum flux density point in the model. Also, all the model has maintained
maximum 1.5 T magnetic flux density in the its all area.
This model has zero power levitation air gap (π§0) is 4.24947 ππ and assume it is 4.3 ππ.
5.1 βπ π¨ and π. π ππ airgap
Figure 16. Flux density in the model β3 π΄ with 1.0 ππ air gap
In this figure shows the color bar and its show the magnetic flux density color values for different
regions.
Top left corner shows the,
β Case number
β Step number and
β Time taken to start given step
Also, down left corner shows the maximum and minimum values of the model
41. 34
5.2 +π π¨ and π. π ππ airgap
Figure 17. Flux density in the model +3 π΄ with 1.0 ππ air gap
5.3 π π¨ and π. π ππ airgap
Figure 18. Flux density in the model 0 π΄ with 1.0 ππ air gap
42. 35
5.4 βπ π¨ and π. π ππ airgap
Figure 19. Flux density in the model β3 π΄ with 4.3 ππ air gap
5.5 +π π¨ and π. π ππ airgap
Figure 20. Flux density in the model +3 π΄ with 4.3 ππ air gap
43. 36
5.6 π π¨ and π. π ππ airgap (zero power levitation air gap (π π))
Figure 21. Flux density in the model 0 π΄ with 4.3 ππ air gap
6 SIMULATION DETAILS
6.1 Mesh
This model has only part mesh elements
β Mesh size = 1.0 ππ
β Air element size = 2.0 ππ
Geometry details
β Number of elements = 1 330 070
β Number of nodes = 235 702
45. 38
7 COMPUTER CHARACTERISTICS
Processor model : Intel(R) coreTM
i5-7200U CPU @ 2.50 GHz
Processer speed
Processer base frequency : 2.50 GHz
Maximum turbo frequency : 3.10 GHz
Number of cores : 2
RAM amount : 8.00 GB
Total space taken to simulation
Size : 63.6 GB (68,290,856,898 bytes)
Size on disk : 63.6 GB (68,291,661,824 bytes)
Contains : 434 Files, 105 Folders
8 LINEARIZE THE EQUATION
For this case, nonlinear equation of maglev model is,
πΉ = π
(π+π)2
(π§+π)2
and,
Constant of a, b, c is
β a = 0.3185
β b = 20.9
β c = 1.149.
In the simulation process, got the π§0 is 0π΄ current with 4.24947 ππ airgap. Then linearize
above equation for
π = π and π = π. πππππ
Rewrite above equation,
πΉ = 0.3185
(π+20.9)2
(π§+1.149)2
(C)
π(π, π§)β0, 4.24947 = π(π0, π§0) +
ππ
ππ
(π0, π§0)(π β π0) +
ππ
ππ§
(π0, π§0)(π§ β π§0)
ππ
ππ
= 2 π₯ 0.3185
(0+20.9)
(4.24947+1.149)2
and
ππ
ππ§
= β2 π₯ 0.3185
(0+20.9)2)
(4.24947+1.149)3
46. 39
Therefore,
π(π, π§)β0, 4.24947 = 0.3185
(0+20.9)2
(4.24947+1.149)2
+ 2 π₯ 0.3185
(0+20.9)
(4.24947+1.149)2
(π β 0) β
2 π₯ 0.3185
(0+20.9)2)
(4.24947+1.149)3
(π§ β 4.24947)
π(π, π§)β0, 4.24947 = 0.3185
(20.9)2
(4.24947+1.149)2
+ 2 π₯ 0.3185
(20.9)
(4.24947+1.149)2
(π) β
2 π₯ 0.3185
(20.9)2)
(4.24947+1.149)3
(π§ β 4.24947)
π(π, π§)β0, 4.24947 = 4.773760475 + 1.768560527(π§ β 4.24947)
π(π, π§)β0, 4.24947 = 1.768560527 π§ + 12.2892053784029
This is the linearize equation for the equation (C). using MATLAB, it can be linearized as shown
in below figure
Figure 22. Linearization of equation (C)
47. 40
9 DISCUSSION
This process had taken more time to finish because of it has simulation process and it took large
time period. in this model has more than 6 hours for each case. Also, other works which related
to the assignment has taken more time. During the finite element analysis (FEA) of the model
time depend on the simulation based on size of model and element size of model which given.
As an example, in here used element size is 1.0 ππ, if used 4.0 ππ it will reduce to 45 mins
per each case. Then time can be reduce increasing the size of the element but it mainly affected
to the output result of the FEA process. Even the shape of graph which we excepted can be
different when the size of elements is higher.
The physical model mainly affects to the result and changing the shape, dimensions of the model
can be design more accurate system. Then we design model from Solid works and test 13, 1,25
cases respectively, to get result which we excepted.
Other thing is saturation of ss400 material. For get accurate result from this model it should be
avoid the saturation. In this report shown in figure 16 and 17 there is no any saturation of the
entire process of this model.
After plotting the 3D curve using MATLAB curve fitting tool it gets 1.263 RMSE value it more
then higher value of excepted one. it can be reduced if we use polynomial function with degree
of 3 for both current and airgap instead given function of equation A, it can be get very closer
value of the excepted value.
Figure 23. cure fitting for J-Mag data with polynomial function
48. 41
Then it given coefficients, RMSE and other details of curve of,
Linear model Poly33:
f(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + p21*x^2*y
+ p12*x*y^2 + p03*y^3
Coefficients (with 95% confidence bounds):
p00 = 48.17 (48.02, 48.32)
p10 = 2.119 (2.079, 2.159)
p01 = -25.74 (-25.9, -25.59)
p20 = 0.466 (0.4559, 0.4762)
p11 = -0.36 (-0.3834, -0.3366)
p02 = 5.026 (4.977, 5.074)
p30 = 0.0009047 (-0.001605, 0.003415)
p21 = -0.05856 (-0.06124, -0.05588)
p12 = 0.01587 (0.01259, 0.01915)
p03 = -0.3407 (-0.3452, -0.3361)
Goodness of fit:
SSE: 54.96
R-square: 0.9993
Adjusted R-square: 0.9993
RMSE: 0.2084
As shown RMSE value is reduced from 1.263 to 0.2084.
49. 42
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