The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) /...

25
The Problem

  • date post

    21-Dec-2015

  • Category

    Documents

  • view

    229

  • download

    3

TAGS:

Transcript of The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) /...

Page 1: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

The Problem

Page 2: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Elements and Nodes

144 in 3

4

5

3

45

192 in 108 in

Plane Truss

2

1

3

1

2

10 kips

100 kips

(AE/L) = (AE/L) = AE/L1 2

Global System Degrees of Freedom

4P , X 4

3P , X 3 5P , X 5

6P , X 6

2P , X 2

1P ,

1 u1

Local Element Degrees of Freedom

X 1

f , u 4 u4f , u

3 u3f , u2 u2f , u

Page 3: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

sin1 = (-12 - 0) / (20) = -0.6cos1 = (16 - 0) / (20) = 0.8sin2 = (12 - 0) / (15) = 0.8cos2 = (9 - 0) / (15) = 0.6

Sines and Cosines

Page 4: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

3 4 1 2

S1 AEL

1

0.64 0.48 0.64 0.48

0.48 0.36 0.48 0.36

0.64 0.48 0.64 0.48

0.48 0.36 0. 48 0.36

3

4

1

2

S2 AEL

2

0.36 0.48 0.36 0.48

0. 48 0.64 0.48 0.64

0.36 0.48 0.36 0.48

0.48 0.64 0.48 0.64

1

2

5

6

1 2 5 6

Element Matrices [S]

Page 5: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

K Sii1

NM

System Stiffness Matrix

Page 6: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

3 4 1 2

S1 AEL

1

0.64 0.48 0.64 0.48

0.48 0.36 0.48 0.36

0.64 0.48 0.64 0.48

0.48 0.36 0. 48 0.36

3

4

1

2

S2 AEL

2

0.36 0.48 0.36 0.48

0. 48 0.64 0.48 0.64

0.36 0.48 0.36 0.48

0.48 0.64 0.48 0.64

1

2

5

6

1 2 5 6

Element Matrices [S]

Page 7: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Summing Element Stiffnesses

.64 -.48 -.64 .48

-.48 .36 .48 -.36

-.64 .48 .64 -.48

.48 -.36 -.48 .36

Page 8: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Summing Element Stiffnesses

.64+.36 -.48+.48 -.64 .48 -.36 -.48

-.48+.48 .36+.64 .48 -.36 -.48 -.64

-.64 .48 .64 -.48

.48 -.36 -.48 .36

-.36 -.48 .36 .48

-.48 -.64 .48 .64

Page 9: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

KAEL

0.64 0.36 0.48 0.48 0.64 0.48 0.36 0.48

0.48 0.48 0.36 0.48 0.48 0.36 0.48 0.64

0.64 0.48 0.64 0.48 0 0

0.48 0.36 0.48 0.36 0 0

0.36 0.48 0 0 0.36 0.48

0.48 0.64 0 0 0.48 0.64

1

2

3

4

5

6

1 2 3 4 5 6

Two Matrix Contributions

Page 10: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

1 2 3 4 5 6

K AEL

1.00 0.00 0.64 0.48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0.64 0.48 0.64 0.48 0 0

0.48 0.36 0.48 0.36 0 0

0.36 0. 48 0 0 0.36 0.48

0.48 0.64 0 0 0.48 0.64

1

2

3

4

5

6

Final [K]

Page 11: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

P1

P2

P3

P4

P5

P6

AE

L

1.00 0.00 0.64 0.48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0.64 0.48 0.64 0.48 0 0

0.48 0.36 0.48 0.36 0 0

0.36 0.48 0 0 0.36 0.48

0.48 0.64 0 0 0.48 0.64

X1

X2

X3

X4

X5

X6

Final Equation P=KX

Page 12: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

System Stiffness Matrices

1

2 4

53

6

7 1

7 6

32

5

4

Node Numbering Scheme 1 Node Numbering Scheme 2

KScheme 1

X X X X X X 0 0 0 0 0 0 0 0

X X X X X X 0 0 0 0 0 0 0 0

X X X X X X X X 0 0 0 0 0 0

X X X X X X X X 0 0 0 0 0 0

X X X X X X X X X X 0 0 0 0

X X X X X X X X X X 0 0 0 0

0 0 X X X X X X X X X X 0 0

0 0 X X X X X X X X X X 0 0

0 0 0 0 X X X X X X X X X X

0 0 0 0 X X X X X X X X X X

0 0 0 0 0 0 X X X X X X X X

0 0 0 0 0 0 X X X X X X X X

0 0 0 0 0 0 0 0 X X X X X X

0 0 0 0 0 0 0 0 X X X X X X

KScheme 2

X X X X 0 0 0 0 0 0 0 0 X X

X X X X 0 0 0 0 0 0 0 0 X X

X X X X X X 0 0 0 0 X X X X

X X X X X X 0 0 0 0 X X X X

0 0 X X X X X X X X 0 0 0 0

0 0 X X X X X X X X 0 0 0 0

0 0 0 0 X X X X X X X X 0 0

0 0 0 0 X X X X X X X X 0 0

0 0 0 0 X X X X X X X X X X

0 0 0 0 X X X X X X X X X X

0 0 X X X X 0 0 X X X X X X

0 0 X X X X 0 0 X X X X X X

X X X X 0 0 0 0 0 0 X X X X

X X X X 0 0 0 0 0 0 X X X X

Page 13: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Solving the System of Equations

Page 14: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Modify for Known Loads

Elements and Nodes

144 in 3

4

5

3

45

192 in 108 in

Plane Truss

2

1

3

1

2

10 kips

100 kips

(AE/L) = (AE/L) = AE/L1 2

Global System Degrees of Freedom

4P , X

4

3P , X3 5P , X5

6P , X

6

2P , X2

1P ,

1 u1

Local Element Degrees of Freedom

X 1

f , u 4 u4f , u

3 u3f , u2 u2f , u

10

100

P3

P4

P5

P6

AE

L

1.00 0.00 0.64 0.48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0.64 0.48 0.64 0.48 0 0

0.48 0.36 0.48 0.36 0 0

0.36 0.48 0 0 0.36 0.48

0. 48 0.64 0 0 0.48 0.64

X1

X2

X3

X4

X5

X6

Page 15: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Modify for Boundary Conditions

Elements and Nodes

144 in 3

4

5

3

45

192 in 108 in

Plane Truss

2

1

3

1

2

10 kips

100 kips

(AE/L) = (AE/L) = AE/L1 2

Global System Degrees of Freedom

4P , X4

3P , X3 5P , X5

6P , X 6

2P , X2

1P ,

1 u1

Local Element Degrees of Freedom

X 1

f , u 4 u4f , u

3 u3f , u2 u2f , u

10

100

P3

P 4

P5

P6

AE

L

1.00 0.00 0.64 0.48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0.64 0.48 0.64 0.48 0 0

0.48 0.36 0.48 0.36 0 0

0.36 0.48 0 0 0.36 0.48

0. 48 0.64 0 0 0.48 0.64

X1

X 2

0

0

0

0

Page 16: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Modify to Ease Solution

10

100

0

0

0

0

AE

L

1.00 0.00 0.64 0.48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0 0 L / AE 0 0 0

0 0 0 L / AE 0 0

0 0 0 0 L / AE 0

0 0 0 0 0 L / AE

X1

X2

X3

X4

X5

X6

Page 17: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Return Symmetry

10 ( 0.64AE / L)(X3) (0.48AE / L)(X4) ( 0.36AE / L)(X5 ) ( 0.48AE / L)(X6 )

100 (0.48AE / L)(X3) ( 0.36AE / L)(X4 ) ( 0.48AE / L)(X5 ) ( 0.64AE / L)(X6 )

0

0

0

0

AE

L

1.00 0.00 0 0 0 0

0.00 1.00 0 0 0 0

0 0 L / AE 0 0 0

0 0 0 L / AE 0 0

0 0 0 0 L / AE 0

0 0 0 0 0 L / AE

X1

X2

X3

X4

X5

X6

Page 18: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Modified Equations

10

100

0

0

0

0

AE

L

1.00 0.00 0 0 0 0

0.00 1.00 0 0 0 0

0 0 L / AE 0 0 0

0 0 0 L / AE 0 0

0 0 0 0 L / AE 0

0 0 0 0 0 L / AE

X1

X2

X3

X4

X5

X6

Page 19: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Recap

Page 20: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Initial Matrix

0

0

0

0

0

0

AE

L

1.00 0.00 0.64 0.48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0.64 0.48 0.64 0.48 0 0

0.48 0.36 0.48 0.36 0 0

0.36 0.48 0 0 0.36 0.48

0. 48 0.64 0 0 0.48 0.64

X1

X2

X3

X4

X5

X6

Page 21: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Loads

10

100

0

0

0

0

AE

L

1.00 0.00 0.64 0.48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0.64 0.48 0.64 0.48 0 0

0.48 0.36 0.48 0.36 0 0

0.36 0.48 0 0 0.36 0.48

0.48 0.64 0 0 0. 48 0.64

X1

X2

X3

X4

X5

X6

Page 22: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Boundary Conditions

10

100

0

0

0

0

AE

L

1.00 0.00 0.64 0. 48 0.36 0.48

0.00 1.00 0.48 0.36 0.48 0.64

0 0 L / AE 0 0 0

0 0 0 L / AE 0 0

0 0 0 0 L /AE 0

0 0 0 0 0 L / AE

X1

X2

X3

X4

X5

X6

Page 23: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Symmetry

10

100

0

0

0

0

AE

L

1.00 0.00 0 0 0 0

0.00 1.00 0 0 0 0

0 0 L / AE 0 0 0

0 0 0 L / AE 0 0

0 0 0 0 L / AE 0

0 0 0 0 0 L / AE

X1

X2

X3

X4

X5

X6

Page 24: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Solution

10 = AE/L X1100 = AE/L X2 0 = AE/L X3 0 = AE/L X4 0 = AE/L X5 0 = AE/L X6

X

10LAE

100LAE

0

0

0

0

Page 25: The Problem. sin 1 = (-12 - 0) / (20) = -0.6 cos 1 = (16 - 0) / (20) = 0.8 sin 2 = (12 - 0) / (15) = 0.8 cos 2 = (9 - 0) / (15) = 0.6 Sines.

Force Calculation (f=sbX)

f i (EAT) i

1

1

AE

L i

1 1

1 1

cosi sin i 0 0

0 0 cosi sini

X1i

X2i

X3i

X4i

t i AEL

iX3i X1i cosi X4i X2i sin i (EA)iT

{(X3i-X1i) cosi + (X4v-X2v) sini} is simply the change in length

t1 = AE/L {(10L/AE - 0)(0.8) + (100L/AE - 0)(-0.6)} + (0)t2 = AE/L {(0 - 10L/AE - 0)(0.6) + (0 - 100L/AE - 0)(0.8)} + (0)

t1 = f2 = -f1 = -52 kipst2 = f4 = -f3 = -86 kips


Linear Second-Order Equations...34 CHAPTER 4. LINEAR SECOND-ORDER EQUATIONS −0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 t y (a) undamped

Linear Second-Order Equations...34 CHAPTER 4. LINEAR SECOND-ORDER EQUATIONS −0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 t y (a) undamped

2-Wire Galvanically Isolated IC Temperature Sensor … Bath Temperature (°C) Temperature Accuracy (°C)-100 -50 0 50 100 150 200-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 D004 TIDA-00752

2-Wire Galvanically Isolated IC Temperature Sensor … Bath Temperature (°C) Temperature Accuracy (°C)-100 -50 0 50 100 150 200-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 D004 TIDA-00752

ROSAEC Center - 분석기의진행율예측하기rosaec.snu.ac.kr/meet/file/20140730m.pdf · 2018. 4. 12. · proftpd-1.3.2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 actual progress

ROSAEC Center - 분석기의진행율예측하기rosaec.snu.ac.kr/meet/file/20140730m.pdf · 2018. 4. 12. · proftpd-1.3.2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 actual progress

Welcome | Cowles Foundation for Research in Economics · 2020-01-03 · 0 100 200 300 400 0 0.2 0.4 0.6 0.8 1 Asset level % 6 0 100 200 300 400 0 0.2 0.4 0.6 0.8 1 Asset level % 6

Welcome | Cowles Foundation for Research in Economics · 2020-01-03 · 0 100 200 300 400 0 0.2 0.4 0.6 0.8 1 Asset level % 6 0 100 200 300 400 0 0.2 0.4 0.6 0.8 1 Asset level % 6

National Climate and Water Briefing - Bureau of Meteorology · Bureau of Meteorology National Climate and Water Briefing 29 September 2015-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 1 Jan

National Climate and Water Briefing - Bureau of Meteorology · Bureau of Meteorology National Climate and Water Briefing 29 September 2015-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 1 Jan

1 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 ......Differencing: Nonseasonal Order=1, Seasonal Order=1 Sample Autocorrelations of the Residuals 【鉱工業 】1-1.0 -0.8 -0.6

1 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 ......Differencing: Nonseasonal Order=1, Seasonal Order=1 Sample Autocorrelations of the Residuals 【鉱工業 】1-1.0 -0.8 -0.6

Presentations can be viewed An advantage of using slides ... · 0 2 4 6 8 10 12-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 time(t) magnitude Before PPF filter After filter 22 Slides should

Presentations can be viewed An advantage of using slides ... · 0 2 4 6 8 10 12-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 time(t) magnitude Before PPF filter After filter 22 Slides should

Keysight E8663D PSG RF Analog Signal Generator · 2019-12-04 · 250 kHz to 2 GHz 6 ± 0.8 ± 0.8 ± 0.6 ± 0.6 ± 0.6 ± 0.7 ± 0.8 > 2 GHz to 9 GHz ± 1.0 ± 0.8 ± 0.8 ± 0.8 ±

Keysight E8663D PSG RF Analog Signal Generator · 2019-12-04 · 250 kHz to 2 GHz 6 ± 0.8 ± 0.8 ± 0.6 ± 0.6 ± 0.6 ± 0.7 ± 0.8 > 2 GHz to 9 GHz ± 1.0 ± 0.8 ± 0.8 ± 0.8 ±

ATALEOFTWOMONSTERS ...hiddenmonsters/talks/robinson.pdfgeometry,$cloud$distribution$etc. 1.0 0.8 0.6 0.4 0.2 0.0 e 0 5 10 15 20 25 Delay 1.8 1.6 1.4 1.2 1.0 0.8 0.6 x 0 5 10 15 20

ATALEOFTWOMONSTERS ...hiddenmonsters/talks/robinson.pdfgeometry,$cloud$distribution$etc. 1.0 0.8 0.6 0.4 0.2 0.0 e 0 5 10 15 20 25 Delay 1.8 1.6 1.4 1.2 1.0 0.8 0.6 x 0 5 10 15 20

PARTII:EVALUATING MANAGEMENTACTIONS€¦ · Volatility of variance 0 0.2 0.4 0.6 0.8 1 Sharpe ratio 0.55 0.6 0.65 0.7 0.75 0.8 Sharpe ratio vs volatility of variance VaR without Vol

PARTII:EVALUATING MANAGEMENTACTIONS€¦ · Volatility of variance 0 0.2 0.4 0.6 0.8 1 Sharpe ratio 0.55 0.6 0.65 0.7 0.75 0.8 Sharpe ratio vs volatility of variance VaR without Vol

Benefiting from Negative Curvature - Northwestern University · k = v k 0 5 10 15 20 25 30 35-1-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 1 heart8ls biggs6 heart6ls engval2 eckerle4ls osborneb

Benefiting from Negative Curvature - Northwestern University · k = v k 0 5 10 15 20 25 30 35-1-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 1 heart8ls biggs6 heart6ls engval2 eckerle4ls osborneb

Robotics TOOLBOX - Lanka Education and Research Network · Robotics TOOLBOX for MATLAB (Release 6) −1 −0.5 0 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8

Robotics TOOLBOX - Lanka Education and Research Network · Robotics TOOLBOX for MATLAB (Release 6) −1 −0.5 0 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8

TP3“ModellierungundHomogenisierungmagneto ... · 2018. 6. 4. · −0.4 −0.2 0 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 −0.8 −0.6 −0.4 −0.2 0 −0.6 −0.4 −0.2 0 0.2 0.4

TP3“ModellierungundHomogenisierungmagneto ... · 2018. 6. 4. · −0.4 −0.2 0 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 −0.8 −0.6 −0.4 −0.2 0 −0.6 −0.4 −0.2 0 0.2 0.4

1.0 Fraction of mass identified 0 - ACP - Recent1.0 0.8 0.6 0.4 0.2 0.0 Fraction of mass identified 1.0 0.8 0.6 0.4 0.2 0.0 Cumulative fraction identified 0 40 80 120 Number of VOC

1.0 Fraction of mass identified 0 - ACP - Recent1.0 0.8 0.6 0.4 0.2 0.0 Fraction of mass identified 1.0 0.8 0.6 0.4 0.2 0.0 Cumulative fraction identified 0 40 80 120 Number of VOC

0201 BasicLine Mono 2014-02 GB RZ · for restricted installation conditions ... Use our free project planning service. ... 0 0.2 0.4 0.6 0.8 0.2 0.4 0.6 0.8

0201 BasicLine Mono 2014-02 GB RZ · for restricted installation conditions ... Use our free project planning service. ... 0 0.2 0.4 0.6 0.8 0.2 0.4 0.6 0.8

15-826: Multimedia Databases and Data Mining€¦ · Galaxies (Sloan Digital Sky Survey w/ B. Nichol)-1-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 1-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8

15-826: Multimedia Databases and Data Mining€¦ · Galaxies (Sloan Digital Sky Survey w/ B. Nichol)-1-0.8-0.6-0.4-0.2 0 0.2 0.4 0.6 0.8 1-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8

1.0—1.65 0.8—1.0 0.6—0.8 0.4—0.6 0.2—0.4 0.0—0.2 -0.95—0 · Title: 05 Created Date: 5/8/2014 7:34:13 PM

1.0—1.65 0.8—1.0 0.6—0.8 0.4—0.6 0.2—0.4 0.0—0.2 -0.95—0 · Title: 05 Created Date: 5/8/2014 7:34:13 PM

Music and Engineering: Musical Instrument Synthesis · 2018. 9. 12. · AMP FREQ PHASE DUR DECAY TIME RISE TIME 0 0.5 1 0 0.2 0.4 0.6 0.8 1 0 0.5 1 0 0.2 0.4 0.6 0.8 1 0 0.5 1-100-80-60-40-20

Music and Engineering: Musical Instrument Synthesis · 2018. 9. 12. · AMP FREQ PHASE DUR DECAY TIME RISE TIME 0 0.5 1 0 0.2 0.4 0.6 0.8 1 0 0.5 1 0 0.2 0.4 0.6 0.8 1 0 0.5 1-100-80-60-40-20

Spectral Weighting Functions for the Effects of Solar ......(gC gChl-1 h-1) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 500 1000 1500 Photosynthesis (gC gChl-1 h-1) 0 0.2 0.4 0.6 0.8 1 1.2 0

Spectral Weighting Functions for the Effects of Solar ......(gC gChl-1 h-1) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 500 1000 1500 Photosynthesis (gC gChl-1 h-1) 0 0.2 0.4 0.6 0.8 1 1.2 0

MATH 590: Meshfree Methodsfass/590/notes/Notes590_Ch1Part3Print.pdf · 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 h X x y Figure:The fill distance for N = 25 Halton points (h X; ˇ0:2667).

MATH 590: Meshfree Methodsfass/590/notes/Notes590_Ch1Part3Print.pdf · 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 h X x y Figure:The fill distance for N = 25 Halton points (h X; ˇ0:2667).