Analysis of Impact Force Equations

Analysis of Impact Force Equations

Prepared for the International Technical Rescue Symposium,

November 2002

By Chuck Weber

Abstract

This paper compares the actual impact forces measured during controlled testing to the values

calculated by a commonly accepted rope force-predicting equation. All tests studied were

simple ¡°free-fall¡± drops where one end of the rope was secured to a rigid anchor and the other

tied to the test weight ¨C no belay devices, rope slippage at anchor or belay device, intermediate

anchors, friction, etc. to factor in.

Through this study it can be shown that the universally accepted manner of modeling life-safety

ropes as springs (along with the corresponding assumptions needed to allow use of the same

equations) is quite accurate when the test weights and fall factors are relatively low. However,

the force predictions of the equation become noticeably less accurate as the test weights and

fall factors increase ¨C as much as 30% low in certain cases.

Data Sets

The two data sets to which this paper

applied the force-predicting equation were

taken directly from recent past ITRS reports:

1) Fall Factors and Life Safety Ropes: a

closer look, ITRS 2001

2) UIAA Dynamic Rope drop testing results

with loads greater than 80 kg, ITRS 1999.

Details on how those tests were conducted

are explained in those papers, but not here.

Force-predicting equation

This paper also does not go into any great

detail about all the assumptions and

principles used to create force-predicting

equations. However, for anyone interested

in further reading or study, Steve Attaway¡¯s

1996 paper (1) does a particularly thorough

and good job of explaining this issue.

Generally speaking, all 4 sources reviewed

and referenced at the end of this paper

used basically the same assumptions and

principles to end up with a force-predicting

equation that is equal to this form when the

needed conversion of units or terms is

applied:

Pigeon Mountain Industries, Inc.

ITRS 2002

Where F = Force (in pounds of force) that

will be generated when these 4 known

values are plugged into the equation:

W = weight (of falling object ¨C in pounds)

h = height of fall (in feet)

M = modulus of rope (in pounds)

L = total length of rope (in feet)

The W, h, and L values are basically selfexplanatory and independent of the type of

rope used, so let¡¯s look at how different

values of M affect the formula¡¯s outcome.

The columns in the left half of the following

table are taken from test paper #1 for all the

drop tests made on PMI 12.mm Classic

Static rope. The columns of the right half

show the results when two different values

are used for M.

Page 1 of 7

31

500

32 (227 kg)

33

34

146

0.25

0.25

0.25

0.25

0.25

1

2

4

5

5

4

8

16

20

20

2285

2663

3077

3131

2917

x

x

x

x

x

33.6%

39.2%

45.3%

46.0%

42.9%

2500

2500

2500

2500

2500

1.094

0.939

0.812

0.798

0.857

93

176

94 (80 kg)

153

96

97

132

0.5

0.5

0.5

0.5

0.5

0.5

1

2

4

8

10

10

2

4

8

16

20

20

1091

1413

1788

2007

2180

2046

x

x

x

x

x

x

16.0%

20.8%

26.3%

29.5%

32.1%

30.1%

1810

1810

1810

1810

1810

1810

1.659

1.281

1.012

0.902

0.830

0.885

35

500

36 (227 kg)

37

38

39

152

0.5

0.5

0.5

0.5

0.5

0.5

1

2

4

8

10

10

2

4

8

16

20

20

2741

3248

4235

5000

5126

5045

x

x

x

x

x

x

40.3%

47.8%

62.3%

73.5%

75.4%

74.2%

3284

3284

3284

3284

3284

3284

1.198

1.011

0.775

0.657

0.641

0.641

98

176

99 (80 kg)

100

101

133

1.0

1.0

1.0

1.0

1.0

2

5

10

20

20

2

5

10

20

20

1633

2358

2961

3426

3176

x

x

x

x

x

24.0%

34.7%

43.5%

50.4%

46.7%

2481

2481

2481

2481

2481

1.519

1.052

0.838

0.724

0.781

40

500

41 (227 kg)

42

1.0

1.0

1.0

2

5

10

2

5

10

3908

5774

6136

x

x

57.5%

84.9%

90.2%

4405

4405

4405

1.127

0.763

0.718

102

176

103 (80 kg)

104

105

2.0

2.0

2.0

2.0

4

8

12

16

2

4

6

8

2587

3632

4434

4697

x

x

x

x

38.0%

53.4%

65.2%

69.1%

3430

3430

3430

3430

1.326

0.944

0.774

0.730

43

500

44 (227 kg)

2.0

2.0

4

8

2

4

6382

6431

x

93.9%

94.6%

6000

6000

0.940

0.933

x

x

Accurate predictions

If you only looked at the first two groups of

0.25 fall factor in the M = 15000 section,

you would likely conclude that the formula is

¡°reasonably accurate¡± (+/- 10%) given the

ratios shown.

The M = 15000 was determined by dividing

300 # by the rope¡¯s elongation at that

weight. The M = 40000 was determined by

dividing the strength of the rope at a very

Pigeon Mountain Industries, Inc.

ITRS 2002

X

X

X

X

X

X

X

X

X

X

X

X

ratio

to actual

2.092

1.718

1.464

1.490

1.595

3702

3702

3702

3702

3702

1.620

1.390

1.203

1.182

1.269

2835

2835

2835

2835

2835

2835

2.599

2.006

1.586

1.413

1.301

1.386

5000

5000

5000

5000

5000

5000

1.824

1.539

1.181

1.000

0.975

0.991

3932

3932

3932

3932

3932

2.408

1.668

1.328

1.148

1.238

6844

6844

6844

1.751

1.185

1.115

5486

5486

5486

5486

2.120

1.510

1.237

1.168

9458

9458

1.482

1.471

X

X

X

high load, just before rope failure, by the

elongation at that point.

Note that the tests with L values of 2, 4, and

5 actually represent an overall very stretchy

section of static rope due to the fact that

most of their length includes the knots at

each connection end. So, it stands to

reason that the force-predicting equation

using an M-value of 15000 along with those

lengths will overstate the actual force

Page 2 of 7

GOOD

(+/- 10%)

FORCE

calc. by

formula

F (lbf)

2060

2060

2060

2060

2060

Rating

EXCELLENT

(+/- 2%)

FORCE

calc. by

formula

ratio

F (lbf)

to actual

1338 1.359

1338 1.116

1338 0.951

1338 0.968

1338 1.036

used M = 40000

Rating

GOOD

(+/- 10%)

drop rope Force

drop test

height length meas.

ref. weight

h

L in test

#

(lb)

FF

(ft)

(ft) F (lbf)

89

176 0.25

1

4

985

90 (80 kg)

0.25

2

8

1199

91

0.25

4

16

1407

92

0.25

5

20

1383

131

0.25

5

20

1292

meas.

Rope

force

Failure? as % of

YES NO knotted

strength

x 14.5%

x 17.6%

x 20.7%

x 20.3%

x 19.0%

EXCELLENT

(+/- 2%)

used M = 15000

PMI 12.5mm Classic Static Rope

recorded during the test. This is another

way to say that the ratio is greater than 1.

If ropes truly acted like springs

The graph below shows the actual slow-pull

test data curves for the three ropes detailed

in this paper.

Another note is that if the calculated F is

greater than the rope¡¯s knotted breaking

strength, that is to say the formula is

predicting rope failure.

Added to this graph are four theoretical

ropes to show what truly ¡°linear¡± forceelongation curves would look like. The Mvalues for these theoretical ropes are:

A = 7500, B = 15000, C = 40000, D = 5000.

You will see that essentially all of the ratios

in the M = 40000 section are overstated.

However, it is interesting to note that this M

value does create very accurate force

predictions when the forces recorded in the

drop test are very high and near the rope¡¯s

breaking strength.

A primary assumption needed to create this

force-predicting equation was that the forceelongation curves of static and low-stretch

ropes were close enough to straight lines

that the ropes¡¯ performance could be

modeled by spring equations. This is what

allows otherwise mathematically complex

equations to be simplified down to the

equation presented.

Drop #42

This particular test makes a convincing

argument that the formula should NOT be

relied upon to predict the force generated in

this fall. For the M = 15000 section the

predicted F value, 4405, is much less

than the knotted rope breaking strength

of 6800, so one would assume that the

rope would NOT fail.

Force-elongation curves for various NEW ropes

and THEORETICAL "CONSTANT" MODULUS ropes

6000

However, it is critical to note that the

rope DID FAIL in this test and the force

recorded at the moment of failure was

6136.

4000

Force (lbf)

The ratio in this example is useful to

look at, but not 100% accurate because

had the rope NOT failed, the force

would have to have been a little higher.

So, we can say that the formula

predicted a force AT LEAST 28%

LOW. Needless to say, this would

surely be unacceptable in anyone¡¯s

book. However, note that the much

higher M value accurately predicts rope

failure for this example.

5000

3000

2000

1000

Other static rope diameters show the

same trends, but are not detailed in this

paper. Overall, this analysis suggests

there is no single M-value that can be

used in the given equation to accurately

predict the force for all scenarios for

static ropes.

Pigeon Mountain Industries, Inc.

0

0.0%

5.0%

10.0%

15.0%

20.0%

25.0%

30.0%

Elongation

PMI 12.5mm Static

Theoretical Rope (B)

Theoretical Rope (D)

ITRS 2002

PMI 13mm LS

Theoretical Rope (A)

PMI 10.6mm dyn.

Theoretical Rope (C)

Page 3 of 7

35.0%

Drop # 70

In a similar fashion to the drop #42 section

before, using M = 8000 (about ? that of the

static rope) for this rope in the formula gives

a force well below the rope¡¯s known knotted

breaking strength (4900 #). But, the rope

actually FAILED in this setup and the force

recorded at the moment of failure was

4688#. Again, the formula predicted a force

that was too low.

X

X

X

X

45

500 0.25

46 (227 kg)

47

48

149

1

2

4

5

5

4

8

16

20

20

1943

2307

2519

2595

2612

x

x

x

x

x

2000

2000

2000

2000

2000

1.029

0.867

0.794

0.771

0.766

76

176 0.50

77 (80 kg)

78

79

80

141

1

2

4

8

10

10

2

4

8

16

20

20

1062

1399

1559

1742

1819

1646

x

x

x

x

x

x

1376

1376

1376

1376

1376

1376

1.295

0.983

0.882

0.790

0.756

0.836

49

500 0.50

50 (227 kg)

51

52

53

1

2

4

8

10

2

4

8

16

20

2360

2964

3704

4042

4197

x

x

x

x

x

2562

2562

2562

2562

2562

1.085

0.864

0.692

0.634

0.610

81

176 1.00

82 (80 kg)

83

84

142

1

5

10

20

20

1

5

10

20

20

1551

2151

2682

2901

2605

x

x

x

x

x

1863

1863

1863

1863

1863

1.201

0.866

0.695

0.642

0.715

68

500 1.00

69 (227 kg)

70

2

5

10

2

5

10

3716

4966

4688

x

x

3372

3372

3372

0.908

0.679

0.719

85

176 2.00

86 (80 kg)

87

88

4

8

12

16

2

4

6

8

2515

3367

3846

4138

x

x

x

x

2556

2556

2556

2556

1.016

0.759

0.665

0.618

X

71

4

2

4584

4531

0.988

X

500 2.00

(227 kg)

Pigeon Mountain Industries, Inc.

x

x

ITRS 2002

X

X

X

X

FORCE

calc. by

formula ratio

F (lbf) to actual

1547 1.780

1547 1.572

1547 1.455

1547 1.419

1547 1.450

GOOD

(+/- 10%)

Rope

FORCE

Failure? calc. by

YES NO formula ratio

F (lbf) to actual

1033 1.189

x

1033 1.050

x

1033 0.972

x

1033 0.948

x

1033 0.968

x

used M = 15000

GOOD

(+/- 10%)

drop rope Force

drop

height length meas.

ref. weight

h

L in test

#

(ft)

(ft) F (lbf)

(lb)

FF

72

4

869

176 0.25

1

73 (80 kg)

8

984

2

74

16

1063

4

75

20

1090

5

140

20

1067

5

used M = 8000

EXCELLENT

(+/- 2%)

Blue Water II +Plus 7/16" (11.6mm) Low-Stretch

EXCELLENT

(+/- 2%)

Apply same analysis to a Low-Stretch

design rope

The table below applies the same method of

analysis to a different type of rope. Many of

the same trends are noticed. The M = 8000

is derived from a 300 # load divided by the

elongation at that load.

2845

2845

2845

2845

2845

1.464

1.233

1.129

1.096

1.089

2107

2107

2107

2107

2107

2107

1.984

1.506

1.351

1.209

1.158

1.280

3779

3779

3779

3779

3779

1.601

1.275

1.020

0.935

0.900

2901

2901

2901

2901

2901

1.870

1.348

1.081

1.000

1.113

5110

5110

5110

1.375

1.029

1.090

X

X

4025

4025

4025

4025

1.600

1.195

1.047

0.973

X

X

7000

1.527

X

X

X

X

X

X

X

Page 4 of 7

Analysis of Impact Force Equations

M = 2200 corresponds to the

test weight, 176#, divided by its

static elongation, 8%.

None of the ratios for this set

were even within 10% of the

actual recorded value. The

formula predicted too low a

value in all cases, but none

resulted in rope failure.

M = 5250 corresponds to the

rope¡¯s impact force during the

first drop test, 1866#, divided

by an approximate maximum

dynamic elongation measured

during that drop, 35.5%.

The ratio column in the table

clearly shows that this M-value

produced many ¡°reasonably

accurate¡± results for a variety

of test weights and drop

heights.

However, it should also be

pointed out that the lowest FF

shown in this table is 0.7 and it

is suspected that had more

tests been conducted for each

test weight group, this high M

value would have produced

less accurate results as the FF

decreased further.

PMI 10.5mm Dynamic Rope

L = 8.5 ft in all tests

Drop

FF

ref. #

Test

Weight

(lb)

M = 5250

Force

meas. Force

ratio to

(lbf) calc.

F (lbf)

actual

h (ft)

15.6 1866

1380 0.739

15.6 2038

1485 0.729

Force

calc.

F (lbf)

2024

2171

ratio to

actual

1.085

1.065

1

2

1.7

1.7

176

200

3

4

5

1.6

1.5

1.4

200

200

200

14.7 1978

13.8 1908

12.9 1830

1448

1409

1369

0.732

0.738

0.748

2113 1.068

2053 1.076

1991 1.088

6

7

8

9

10

11

1.7

1.6

1.5

1.4

1.3

1.2

225

225

225

225

225

225

15.6

14.7

13.8

12.9

11.9

11.0

2261

2190

2136

2048

1925

1854

1590

1550

1509

1467

1424

1378

0.703

0.708

0.707

0.716

0.740

0.743

2317

2255

2191

2126

2058

1987

1.025

1.030

1.026

1.038

1.069

1.072

X

12

13

14

15

16

17

18

19

1.7

1.6

1.5

1.4

1.3

1.2

1.1

1.0

250

250

250

250

250

250

250

250

15.6

14.7

13.8

12.9

11.9

11.0

10.1

9.2

2499

2383

2315

2190

2128

2039

1900

1842

1691

1649

1606

1562

1516

1468

1419

1367

0.677

0.692

0.694

0.713

0.712

0.720

0.747

0.742

2456

2391

2324

2255

2183

2109

2031

1950

0.983

1.004

1.004

1.030

1.026

1.034

1.069

1.059

X

X

X

20

21

22

23

24

25

26

27

28

1.7

1.6

1.5

1.4

1.3

1.2

1.1

1.0

0.9

276

276

276

276

276

276

276

276

276

15.7

14.8

13.9

13.0

12.0

11.1

10.2

9.3

8.4

2740

2632

2549

2383

2297

2167

2051

1915

1816

1797

1753

1708

1662

1614

1564

1512

1458

1401

0.656

0.666

0.670

0.697

0.703

0.722

0.737

0.761

0.772

2603

2535

2464

2392

2317

2239

2158

2073

1984

0.950

0.963

0.967

1.004

1.009

1.033

1.052

1.082

1.092

29

30

31

32

33

34

35

36

37

38

39

1.7

1.6

1.5

1.4

1.3

1.2

1.1

1.0

0.9

0.8

0.7

301

301

301

301

301

301

301

301

301

301

301

15.7

14.8

13.9

13.0

12.0

11.1

10.2

9.3

8.4

7.4

6.5

3046

2793

2686

2575

2460

2368

2312

2175

2052

1895

1768

1891

1846

1799

1751

1701

1649

1595

1538

1479

1417

1351

0.621

0.661

0.670

0.680

0.691

0.696

0.690

0.707

0.721

0.748

0.764

2732

2661

2588

2512

2434

2353

2268

2179

2086

1988

1884

0.897

0.953

0.963

0.976

0.989

0.993

0.981

1.002

1.017

1.049

1.066

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

It can be seen in the graph that

the curve for dynamic rope does not curve upward as quickly as the other two shown. Instead,

for any given force the dynamic rope elongates more (shown by curve stretching to the right), as

is expected, than the others. This is what is meant by it has a lower modulus.

Pigeon Mountain Industries, Inc.

Page 5 of 7

GOOD

(+/- 10%)

M = 2200

EXCELLENT

(+/- 2%)

Dynamic Ropes

Using the data set from paper

#2, the following table was

created in a similar fashion as

before. Again both low and

high moduli were used to

compare what force the

equation will predict for very

different M-values.

ITRS 2002

X

X

 ................
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