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EFFECT OF GRADE ON KINETIC FRICTION IN ACCIDENT RECONSTRUCTION

Copyright (C) George M Bonnett, J.D. 1992

ABSTRACT

This paper details a method of adjusting the drag factor of a vehicle when there is a difference of grade between the tested surface and the actual accident surface. This method is based on the sum of the forces acting on the object and is accurate for any grade. The popular method used by most accident reconstructionists consists of adding the signed difference in grade to the test drag factor. Many reconstructionists believe that the adjusted drag factor obtained by this second method is relatively accurate only for grades of ten percent (.1) or less.

THE RECONSTRUCTION OF AN ACCIDENT often involves taking measurements that are difficult or dangerously impractical for the investigator. One of these measurements is the drag factor of the vehicle (or object in question) at the scene of the accident. Often the actual vehicle is not available for the tests and it is not always possible to perform the tests at the scene of the accident. There may be times when the slope of the roadway makes the testing unreasonably dangerous, placing the investigator in the same situation as the victim. As a result, it is often necessary to test a similar surface in an area not exposed to the dangers of the accident scene. This test area may not always be on the same grade as the accident scene and the reconstructionist must adjust for the difference.

The major schools of accident reconstruction generally gloss over the explanation of how to adjust the drag factor for grade, teaching that the acceptable method is to add the signed value of the difference in the percent of grade to the coefficient of kinetic friction (often referred to as drag factor). When the correct method is discussed, it is generally done so in a cursory manner and soon forgotten by the student. This concept pales in comparison to the shear volume of information presented in most courses on accident reconstruction since most of the road surfaces the reconstructionist is likely to encounter have relatively shallow slopes of less than fifteen percent (.15) and the popular method of adjustment will usually suffice. If the slope is greater than ten percent (.1) the error starts to become unacceptable and an alternate method should be used. The strength of the popular method is its simplicity but credence must also be given to those who believe that even though it may be impossible to obtain 100 percent accuracy in every aspect of the reconstruction, every reasonable effort should be made to maintain accuracy. Regardless of the method used, it is beneficial to understand how and why the adjustment should be accomplished.

Gravity effects the coefficient of kinetic friction when it is adjusted for grade in two different ways. The first is in a direction perpendicular to the road surface at the changed grade. This effect can be determined by multiplying by the cosine of the angle of grade (or the cosine of the arctangent of the percent of grade - cos[atn[m]]). The second is in a direction parallel to the surface of the roadway at the changed grade. This effect can be determined by multiplying by the sine of the angle (sine of the arctangent of the percent of grade - sin[atn[m]]). For a positive change of grade this force will act in the same direction as the friction force and therefore must be added to the friction force. The following derivations may assist in understanding this principle

DERIVATIONS

E F - sum of the forces acting on an object

E Fx - the sum of the forces on the X axis

E Fy - the sum of the forces on the Y axis

f - coefficient of kinetic friction over a level surface

F - drag factor - the coefficient of kinetic friction (f) adjusted for [braking and/or] grade (m). A negative acceleration factor

fr - friction force

g - the rate of acceleration due to gravity

G - a pure number obtained by dividing the force of gravity by 32.2 ft/sec^2

N - normal force

W - weight

1. E Fx = fr + m g sin O = m ax
2. E Fy = +N - m g cos O = 0
3. fr = f N
4. E Fy = f N - f m g cos O = 0 …………………………………………….[multiply eq. 2 by f]
5. E Fy = fr - f m g cos O = 0 ………………………………………………[substitute eq. 3 into eq. 4]
6. E Fy = f m g cos O - fr = 0 ………………………………………………[multiply eq. 5 by -1]
7. E F = f m g cos O + m g sin O = m ax …………………………………..[add eq. 1 to eq. 5]
8. E F = m ax = m g (f cos O + sin O) ……………………………………...[re-arrange terms]
9. ax = g(f cos O + sin O) …………………………………………………..[divide eq. 8 by m]
10. ax = g(f cos O + sin O) …………………………………………………..[acceleration on the X axis]
11. F = a/g …………………….……….....[acceleration factor (F) equals acceleration (a) divided by acceleration rate of gravity (g)]
12. O = atn m ………………………….....[angle = arctangent of the % of grade (m)]
13. F = f cos O + sin O ……………….....[divide eq. 10 by g]
14. F = sin(atn m) + f cos(atn m) ……....[substitute atn m for O in eq. 13]

The adjusted drag factor (F) for a positive change of grade, is a combination of both the coefficient of kinetic friction (f) of the object and of the force of gravity or "G" force acting on the object. The coefficient of kinetic friction (f) modified by a positive grade has the force of gravity assisting in the acceleration of the object. Both of these forces combined result in what the reconstructionist refers to as the adjusted drag factor (F).

The following table shows the actual adjusted drag factor (F ACTUAL) along with the drag factor produced with the (f + m) method of approximation. As can be readily observed, the error becomes significant for grades in excess of ten percent (.1). The table also demonstrates that the "drag factor" is not the same for both a positive and negative grade of the same slope.

COEFFICIENT OF KINETIC FRICTION ADJUSTED FOR GRADE

 % GRADE ---------------- ANGLE --------------- f + m---------------------- F-ACTUAL --------------- % ERROR VERTICAL 90.0000 INFINITY 1.0000 INFINITY 10000 89.4271 101.0000 1.0099 9,900.5000 1000 84.2894 11.0000 1.0945 904.9876 500 78.6901 6.0000 1.1767 409.9020 200 63.4349 3.0000 1.3416 123.6068 100 45.0000 2.0000 1.4142 41.4241 90 41.9872 1.9000 1.4123 34.5362 80 38.6598 1.8000 1.4056 28.0625 70 34.9920 1.7000 1.3927 22.0656 60 30.9638 1.6000 1.3720 16.6190 50 26.5651 1.5000 1.3416 11.8034 40 21.8014 1.4000 1.2999 7.7033 30 16.6992 1.3000 1.2452 4.4031 20 11.3099 1.2000 1.1767 1.9804 10 5.7106 1.1000 1.0945 0.4988 LEVEL 0.0000 1.0000 1.0000 0.0000 -10 -5.7106 0.9000 0.8955 0.4988 -20 -11.3099 0.8000 0.7845 1.9804 -30 -16.6992 0.7000 0.6705 4.4031 -40 -21.8014 0.6000 0.5571 7.7033 -50 -26.5651 0.5000 0.4472 11.8034 -60 -30.9638 0.4000 0.3430 16.6190 -70 -34.9920 0.3000 0.2458 22.0656 -80 -38.6598 0.2000 0.1562 28.0625 -90 -41.9872 0.1000 0.0743 34.5362 -100 -45.000 0.0000 0.0000 0.0000 -200 -63.4349 -1.0000 -0.4472 123.6068 -500 -78.6901 -4.0000 -0.7845 409.9020 -1000 -84.2894 -9.000 -0.8955 904.9876 -10000 -89.4271 -99.0000 -0.9900 9,900.5000 -INFINITY -90.000 -INFINITY -1.0000 INFINITY

The 90 degree vertical position in the above table shows an actual F of 1.0000. This is the force of gravity or "G" force that is accelerating (decelerating) the object as it travels in a positive vertical direction. When the vehicle is level the coefficient of kinetic friction is 1.0000 and therefore the "drag factor" is 1.0000 by either method of computation. The force of gravity is perpendicular to the road surface. The -90 degree vertical position in the table shows an actual F of -1.0000. The surface of the roadway has no effect on the vehicle and gravity is the only force acting upon it. The vehicle is in free fall as no force is acting in opposition to gravity in order to decelerate the vehicle. Since the acceleration factor would be 1.0000 for this condition, the deceleration or negative acceleration factor is -1.0000.

In the determination of the forces acting on a vehicle, the reconstructionist must take into consideration the acceleration mentioned above. This force acts to accelerate the vehicle at any negative angle just as it acts to accelerate (decelerate) the vehicle at any positive angle. The magnitude of this force is obtained by multiplying the acceleration factor by the sine of the negative angle and by the acceleration rate due to gravity of -32.2 ft/sec^2. Therefore, given a level coefficient of friction of 1.0000, a vehicle on a downhill slope of 45 degrees would have an adjusted acceleration or drag factor of zero (0.0000) and a vehicle going downhill vertically would have an acceleration factor of 1.0000 or a drag factor of -1.0000.

It is sometimes necessary to obtain the coefficient of friction for a level surface (f) when the test drag factor (F) is known for the grade. The derivations for the equations used to adjust the test drag factor (F) to level grade are shown below.

15. F - sin O = f cos O ………………………………….[subtract sin O from eq. 13]

16. f = (F - sin O) / cos O ………………………………[divide by cos O and re-arrange terms]

17. f = (F - sin(atn m))/cos(atn m) ……………………..[substitute atn m for O in eq. 16]

If the test drag factor (F) is obtained for a surface that is not level, it must first be adjusted to level. It is only after this adjustment to level has been made that the drag factor on grade can be calculated. It would be mathematically incorrect to adjust directly from one non-level grade to another non-level grade as can be seen in the examples below.

EXAMPLES

Example A illustrates the CORRECT method of going from a +.03 test surface with a 1.0000 drag factor to a -.02 grade accident scene:

Ex. A f (level) = (F - sin (atn m))/cos(atn m)

f (level) = (1 - sin (atn .03))/cos(atn .03)

f (level) = .9704

F (@ -.02) = sin (atn m) + f cos (atn m)

F = sin (atn -.02) + .9704 * cos (atn -.02)

F adjusted = .9503

Examples B1 and B2 illustrate two INCORRECT methods of adjusting the drag factor directly for the change of grade (without solving for level), given the identical situation as described above.

Ex. B1 f = (1 - sin (atn (-.05))/cos(atn -.05)

f = 1.0512

Ex. B2 F = sin (atn -.05) + 1 * cos (atn -.05)

F = .9488

Example C illustrates the "f + m" method and the resultant INCORRECT solution to the grade problem in Example A.

Ex. C f = 1 + (-.05)

f = .9500

Example C gives a solution very close to the correct solution obtained in example A. This is a result of the minimal grade used in the examples and an "offsetting error condition" in going from a positive test grade to a negative accident grade.

The equations listed are direction sensitive and must be used with the signed (+/-) value of the angle. The positive grade or angle must be used when dealing with a positive slope, regardless of whether the adjustment is from level to the positive slope (eq. 13 or 14) or from the positive slope to level (eq. 16 or 17). When going from level to a negative slope (eq. 13 or 14) or going from a negative slope to level (eq. 16 or 17) a negatively signed angle must be used. The signing of the angle determines if the force of gravity is assisting or resisting in the deceleration. While this last statement may argue the use of a negative angle if going from a positive slope to level, a distinction must be made between a negative slope (going downhill), and returning to level from a positive slope. It would be mathematically incorrect to use a negative angle when returning to level from a positive slope, or to use a positive angle when returning to level from a negative slope.

In this discussion of the forces acting on a vehicle, air resistance has not been mentioned. While air resistance is always acting on a vehicle, because of the relative short distances and the minimal effect this force has on a vehicle traveling at moderate speeds, and the impossibility of determining the velocity of the wind at the time and place of the actual accident, it's effect has not been included in the computations.

The following computer generated results were obtained while tracking two vehicles simultaneously using the f + m method for unit #1 and the F adjusted method for unit #2. The program (REC-TEC, REC-TEC LLC) uses the method outlined above for computation of the adjusted drag factor but can be used to simulate the f + m method. Both vehicles were tracked in a deceleration from an initial velocity of 100 feet per second to a final velocity of zero (0). The base coefficient of friction was one (1.0000) for both vehicles and a negative grade of twenty percent (-.2) was used in the computations.

TIME/DISTANCE 2

 UNIT #1: DECELERATION ------------------------------------------ UNIT #2: DECELERATION f (TEST) = 1 f (TEST) = 1 GRADE(T) = 0 GRADE(T) = 0 f(LEVEL) = 1 f(LEVEL) = 1 GRADE = -.2 GRADE = -.2 BRAKING = 100 BRAKING = 100 D/FACTOR = .8 D/FACTOR = .7844 RATE OF DECELERATION RATE OF DECELERATION RATE = 25.76 F/S/S RATE = 25.2597 F/S/S RATE = 17.5636 M/H/S RATE = 17.2225 M/H/S DATA (V1 -> V2) DATA (V1 -> V2) DISTANCE = 194.0993 FT DISTANCE = 197.9433 FT TIME = 3.8819 S TIME = 3.9588 S DATA (V1 -> ZERO) DATA (V1 -> ZERO) DISTANCE = 194.0993 FT DISTANCE = 197.9433 FT TIME = 3.8819 S TIME = 3.9588 S INITIAL INITIAL VELOCITY = 100 F/S VELOCITY = 100 F/S VELOCITY = 68.1818 M/H VELOCITY = 68.1818 M/H FINAL FINAL VELOCITY = 0 F/S VELOCITY = 0 F/S VELOCITY = 0 M/H VELOCITY = 0 M/H

 DISTANCE (FT) INTERVAL UNIT #1: 50.0000 UNIT #1 UNIT #2 TIME(S) .................... DIST(FT) .......... VEL(F/S) ................... VEL(M/H) ..... DIST(FT) ..... VEL(F/S) ..... VEL(M/H) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1.9703 50.0000 50.7543 34.6052 49.0290 49.7687 33.9332 2.7864 100.0000 71.7774 48.9392 98.0581 70.3836 47.9888 3.4126 150.0000 87.9090 59.9380 147.0871 86.2019 58.7740 3.8820 194.0994 100.0000 68.1818 190.3301 98.0581 66.8578 DISTANCE (FT) INTERVAL UNIT #2: 50.0000 UNIT #1 UNIT #2 TIME(S) DIST(FT) VEL(F/S) VEL(M/H) DIST(FT) VEL(F/S) VEL(M/H) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1.9897 50.9902 51.2544 34.9462 50.0000 50.2591 34.2676 2.8138 101.9804 72.4847 49.4214 100.0000 71.0771 48.4617 3.4462 152.9706 88.7752 60.5286 150.0000 87.0513 59.3532 3.9589 201.7872 100.0000 68.1818 197.9433 100.0000 68.1818 TIME(S) INTERVAL 0.5000 UNIT #1 UNIT #2 TIME(S) DIST(FT) VEL(F/S) VEL(M/H) DIST(FT) VEL(F/S) VEL(M/H) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.5000 3.2200 12.8800 8.7818 3.1575 12.6299 8.6113 1.0000 12.8800 25.7600 17.5636 12.6299 25.6299 17.2226 1.5000 28.9800 38.6400 26.3455 28.4172 37.8896 25.8338 2.0000 51.5200 51.5200 35.1273 50.5195 50.5195 34.4451 2.5000 80.5000 64.4000 43.9091 78.9367 63.1494 43.0564 3.0000 115.9200 77.2800 52.6909 113.6689 75.7793 51.6677 3.5000 157.7800 90.1600 61.4727 154.7160 88.4092 60.2790 3.8820 194.0994 100.0000 68.1818 190.3301 98.0581 66.8578 3.9589 201.7872 100.0000 68.1818 197.9433 100.0000 68.1818

CONCLUSIONS

The equations presented herein can be utilized for adjusting the coefficient of kinetic friction for grade as applied in accident reconstruction. They present an accurate alternative to the approximation method currently in widespread use. The equations and their derivations are a useful tool for those who use either method.

REFERENCES

George B. Arfken, David F. Griffing, Donald C. Kelley, Joseph Priest , UNIVERSITY PHYSICS
Miami University, Oxford, Ohio, Academia Press, Inc.
Orlando, Florida, 1984

J. Stanndard Baker, TRAFFIC ACCIDENT INVESTIGATION MANUAL
Northwestern University Traffic Institute
Evanston, Illinois, 1975

John Daily, FUNDAMENTALS OF TRAFFIC ACCIDENT RECONSTRUCTION
Institute of Police Technology and Management
Jacksonville, Florida, 1988

Edward R. McCliment, PHYSICS
University of Iowa, Harcourt,Brace Jovanovich, Inc.
Orlando, Florida, 1984

Ross Mocklin, John Rigol, Genovieve May ADVANCED TECHNICAL ACCIDENT INVESTIGATION
Louisiana Department of Public Safety - Office of State Police
Baton Rouge, Louisiana, 1979

Staff of Research and Education Association, THE PHYSICS PROBLEM SOLVER
Research and Education Association
New York, New York, 1983

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