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ACKNOWLEDGEMENT

We would like to thank Dr. L. K. Maheshwari, Vice Chancellor of BITS PILANI for introducing the course Production Techniques for the mechanical students. We are also very grateful to Dr. K. S. Sangwan to give us this opportunity to work in BITS workshop which helped us to materialize our theoretical knowledge into practical one. We would like to thank our instructor Dr. R. P. Mishra who guided us in the practical sessions of Production Techniques and supervised our project. Besides this, we are grateful to the entire staff of workshop for being helpful enough and to give us time from their busy schedule.

ABSTRACT

The presented report details the manufacturing techniques, calculations, and design issues involved in production and fabrication of rack & pinion steering assembly. The rack & pinion steering assembly thus manufactured has a turning radius of 3.5 meters. The assembly is fabricated up to the steering knuckles, which mounted on support stands for demonstration. The report aims at providing precise information about design issues of steering mechanism and techniques of production processes employed.

CONTENTS

I. INTRODUCTION

II. DESIGN ISSUES & CALCULATIONS

III. MANUFACTURING PROCEDURE & CALCULATIONS

IV. REFERENCES

V. APPENDIX A

VI. APPENDIX B: PART-BY-PART CAD MODEL

1 2 4 17 18 23

1

MAN U FAC T U RI N G O F RAC K & PI N I ON S T EERI N G AS S EMB L Y

INTRODUCTION

A rack and pinion is a pair of gears which convert rotational motion into linear motion. The circular pinion engages teeth on a flat bar - the rack. Rotational motion applied to the pinion will cause the rack to move to the side, up to the limit of its travel.

Rack-and-pinion steering is quickly becoming the most common type of steering on cars, small trucks and SUVs. A rack-and-pinion gear set is enclosed in a metal tube, with each end of the rack protruding from the tube. A rod, called a tie rod, connects to each end of the rack.

The pinion gear is attached to the steering shaft. When you turn the steering wheel, the gear spins, moving the rack. The tie rod at each end of the rack connects to the steering arm on the spindle.

The rack-and-pinion gear-set does two things:

• It converts the rotational motion of the steering wheel into the linear motion needed to turn the wheels.

• It provides a gear reduction, making it easier to turn the wheels.

2

The rack -- also known as a steering rack -- is a long piece of metal that is flat on at least one side. The flat side contains teeth running the length of the rack. The teeth are cut perpendicular to the edges of the rack, meaning they run side by side from one end of the rack to the other.

The other major component, the pinion is a round rod that also has teeth on it, although these teeth run parallel to the length of the shaft. The pinion shaft comes into the rack at a ninety-degree angle and the teeth on the pinion mesh with the teeth on the rack. The pinion is connected directly to the steering column, so when the steering wheel is turned to the left, for instance in the case of Ackerman geometry the pinion rotates counter-clockwise .Thus the rotary motion of the pinion is changed to transverse motion by the rack. The rack moves to the right, making the wheels go left. The car turns left.

DESIGN ISSUES & CALCULATIONS

RACK & PINION CALCULATIONS

For pinion,

Diametrical Pitch P = 12 teeth/inch

Number of teeth

Pitch diameter d = N/P = 22/12 = 1.833 inches

Module m = d/N = 1.83/22 = .083 inches

Circular pitch p = dπ/N = 3.14*.083 = 0.262 inches

Addendum = 1/P = 1/12 = 0.083 inches

Dedendum = 1/P = 1/12 = 0.083 inches

OD = 2*addendum + d

OD = 2*0.083 + 1.833 = 2 inches

For rack,

p’ = 0.262 inches

p’ = p*cosα

Here, α is pressure angle

3

p’ = 6.3/22 inches

So (6.3/22)*cosα = 0.262

So, α = 24 degrees

ENGAGEMENT OF RACK AND PINION

Casing inner diameter = 25.4 mm

Rack diameter = 22 mm

Radial clearance = (25.4 – 22)/2 = 1.7 mm

Sleeve thickness = 1mm

PINION AND BEARING HOUSING

The reducer is similar in shape and design to the part of the rack casing in which the pinion and bearing are housed. The reducer was chosen keeping mind the outer diameter of the pinion and bearing used. Consecutively, the length of the shaft that holds the pinion and bearing was ascertained.

SHAFT DESIGN

A shaft was manufactured as per the above design. The length and diameter of different sections of the shaft was determined to press-fit the pinion, bearing and the universal cross joint on it.

STEERING ROD

A hollow tube was used as a steering rod to reduce the weight of the assembly. The ends were made of solid rods which needed proper profiling. On one side it was welded to the steering wheel and on the other end the universal cross joint was mounted.

STEERING KNUCKLE

Length of knuckle = 8.5”

The plate used is 4” wide and .5” thick.

Kingpin Angle or Steering Inclination Angle (SIA) = 12 degrees (To ensure smaller scrub radius and hence more responsive steering)

Caster angle = -7degrees (negative caster ensures more responsive steering)

4

STEERING (ACKERMAN) ARM

The steering arm is 5.14” long and makes an angle of 26.56degrees with the knuckle. These dimensions were chosen in order to get a turning radius of less than 4m for a vehicle of wheelbase 46.5” and track width 53.5”.

Ackerman angle = 26.56degrees

Distance between tie rod ends = 41.9”

Turning Radius = 3.5 meters

*Refer to Appendix A for the calculations

MANUFACTURING PROCEDURE & CALCULATIONS

The rack & pinion steering assembly can be split into smaller components for ease of understanding:

Rack Pinion Casing Bearing & Pinion Housing Steering Knuckle Support stand and steering

rod Ball-joints, Tie-rods Painting and Lubrication

A case-by-case description of each component will be given, explaining its function, the design issues, calculations, along with flow-chart of manufacturing operations on each component.

PINION

One of the crucial components of the assembly, the manufacturing of the pinion requires exhaustive design calculations which have a net effect on the specifications of the entire steering assembly.

The pinion is the driving gear in the arrangement. As discussed earlier, we have decided upon a spur gear arrangement. The pinion’s rotational motion gives translational motion to the driven component that is, the rack.

5

MATERIAL

Mild steel was chosen as the raw material for the pinion, because:

• Mild steel is the most common form of steel • Priced relatively cheaper (about Rs. 40 per kg.) • M.S contains 0.16-0.29 % carbon, hence it gives optimum ductility and good

machinablity

PROCEDURE

BLANK PREPARATION

For the preparation of the blank, some initial calculations were made, which relate to the design issues of the entire assembly.

Size of pinion: Considering one revolution of pinion for covering entire rack-length, we have

inches

Fixing the size of the pinion after considering available rack-length, as D = 2 inch,

inches

Thickness of pinion: Earlier, choice of rod for rack was made as 1 inch thick, 24 inch long.

Hence the thickness of pinion should be less than 1 inch for proper engagement (will be explained later). Thus, thickness of pinion was fixed at 0.5 inch.

Drilling: For mounting of the pinion on the housing, a 10 mm hole had to be drilled.

Processes carried out on the raw material which was cut out using power hack-saw included:

Blank preparation

•Turning, facing, drilling

Teeth cutting

• Involute-gear cutting on milling maching

6

1. Turning on lathe to achieve 2 inch diameter 2. Facing on lathe to achieve 0.5 inch thickness 3. Drilling on lathe with 10 mm drill-bit

TEETH CUTTING

Teeth-cutting was performed by a horizontal milling operation with an involute cutter. Indexing was done with help of dividing head. Following calculations are critical to be made before the operation.

A 12 teeth per inch DP cutter was available. Hence, on a 2 inch diameter blank, number of teeth was found as 22 earlier.

Pitch is given as inches

On the column and knee type milling machine, the blank was mounted in a horizontal milling arrangement. An involute cutter of specifications 12 DP, 21-25 teeth was mounted on the arbor.

Indexing was done with the help of dividing head. For making the equal spaced markings on the blank, the indexing plate had to be set so that exactly 22 teeth would be obtained at the end of this procedure. The gear ratio of the indexing gear to the gear to be manufactured is where N = No. of teeth to be cut = 22.

This gives the number , such that the denominator is number available on the indexing plate. The indexing arm was moved to cover one entire revolution plus 63 holes after every tooth was marked or cut.

The depth of cut was 0.166 inch and was adjusted by raising the table by the required amount before the cutting operation.

Thus a pinion with 22 teeth and 12 teeth per inch was obtained.

RACK

The rack is a straight rod which is driven by the pinion in translational motion. The rack is actually a gear of infinite pitch diameter1

1 Mechanical Engineering Design by Shigley

in a common spur gear arrangement.

7

MATERIAL

As discussed before, Mild Steel is most commonly available steel and easily machinable. Further, as per our design, a 22 mm thick and 24 inch long rod was required. A one inch thick rod was easily available in workshop store.

PROCEDURE

PREPARATION OF ROD

Cutting: Available rod of one inch from the store was cut out to 26 inches (2 inches extra holding in head stock) with the help of power hack-saw.

Turning: It was turned on the lathe for achieving the desired thickness of 22 mm and finished with flat file. Smooth movement of rod inside pipe is very critical, so this operation was essential.

FLAT SURFACE PREPARATION

As previously discussed, a rack of 6.3 inches was to be cut on the rod. For the involute profile to be cut accurately a flat surface was required to be made on this rod. This flat surface was required to be 0.5 inches thick to match with the pinion thickness.

Hence, the depth to be cut can be calculated as

Thus, X is equal to 2mm. This is the depth to be cut.

The shaping was done in four passes of 0.5 mm depth of cut each during pass.

TEETH CUTTING

For appropriate meshing of teeth the pitch of both components has to be equated as

Linear Pitch = Circular Pitch = 0.262 inches

The teeth on the rack were cut using the column and knee type milling machine in the horizontal arrangement. Involute cutter of 12 DP, 21-25 teeth was mounted on the arbor.

Preparation of the rod

•Cutting •Turning

Flat surface preparation

•Shaping

Teeth cutting

•Involute gear cutting on milling machine

Assembly

•Internal thread cutting•Tapping

11

6.35

X

8

In this case indexing was simply done by moving the table by 0.262 inches after every tooth was marked or cut. A depth of cut of 0.166 inches was given by raising the table.

Thus, a length of 6.3 inches resulting in 22 teeth for the above mentioned pitch was obtained.

ASSEMBLY

The rack rod is connected to the ball-joint at the ends through a threaded joint. Ball-joints were obtained from the old SAE BAJA 2009 vehicle part which had a Lotus Elise steering assembly. The ball-joint has an external threaded end which connects to the rack-rod.

For this purpose, an internal thread had to be made on the ends. This was done by drilling, internal threading and tapping, in sequence. Appropriate measurements had to be taken beforehand.

Measurement: The minor diameter was measured with vernier callipers to be 10 mm. Pitch was measure with a metric thread gauge and found to be 1.25 mm. Referring to the fine-pitch, metric threads table2

CASING

, we can obtain the specifications for design purpose (as discussed earlier).

Drilling was done with the help of a 10 mm drill-bit mounted on tail-stock of lathe. Threading was done with a boring tool and automatic feed by engaging the split-nut of arbon with the lead screw on lathe. Tapping was done with a tapping tool of required thread size mentioned above.

This set of operations was performed on each side of the rack-rod. Thus, operations on this component were complete, and the rod was readied for assembly.

Casing is the covering on the rack in which the rack moves to and fro when the steering wheel is turned. It is also used as a carrier of the lubricant for the rack to move smoothly.

MATERIAL

The casing is generally a cast product. But due to limitations and difficulties in casting, a G.I. (Galvanised Iron) pipe of required dimension is used. Galvanized metal has a zinc coating on it to protect it from rust. Further, the inner diameter of the casing is critical as it should be just larger than rack diameter with radial clearance given for lubrication.

Pipe selection: The inner diameter of the casing should just enough to allow the movement of the rack inside itself. We can evaluate the ID of the required pipe as:

Casing inner diameter = rack diameter +2* radial clearance = 22 + 2 x 1.7 = 25.4 mm = 1” 2 Table 8-1, Pg. 398, Shigley, Mechanical Engineering Design

9

PROCEDURE

Cutting to appropriate length: Calculation of appropriate length of pipe was critical and an interesting problem. Consider 6.3 inches of teeth on the rack-rod section. The requirement is that (as shown in figure below), the rack length should be completely exhausted when the last tooth on the corresponding side reaches the pinion. Thus,

Casing length = rack-rod length – rack length = 24 – 6.3 = 17.7 inches.

Slot-making: A wide slot was to be made in the casing through which the pinion engages with the rack. The width of the slot is determined by a minimum width equal to 0.5 inch, which is the thickness of pinion. Beyond this width, the depth the slot from the top becomes critical for engagement of gear-teeth.

Hence, end-milling was done after opening up the casing by drilling holes in its surface. End-milling was performed on the Column and Knee type milling machine, in the vertical milling arrangement. End-milling cutter was used and feed given was along both X and Y coordinates, for obtaining a rectangular slot of 1 inch x 2 inch dimensions.

Joining of Casing and Pinion Housing (Reducer): After the pinion-housing was completed (discussed below) in parallel, both parts were welded together after exhaustive edge-preparation by grinding.

Engagement of Rack & Pinion: For proper engagement of the gear-teeth, generally a spring-lock system is provided just opposite to the pinion, however, since such a system could not be fabricated, two semi-circular sleeves were provided below the rack, to push it onto the pinion teeth. This pair of sleeves (on either side, on each end) were gas-welded with brass as filler material.

Cutting to appropriate length

•Power hack-saw

Slot making

•Drilling and End milling

Joining with pinion housing

•Grinding•Arc-welding

Engagement of gear-teeth

•Sleeve-making•Gas-Welding

6.3”

10

BEARING AND PINION HOUSING

The pinion is engaged on the rack inside a housing arrangement. This housing also supports a bearing behind the pinion. The pinion and bearing are mounted on a shaft, which connects to the Universal cross joint. On the other side of this joint, is the steering rod. The outer-case of the housing assembly was welded to the pipe-casing.

MATERIAL

The entire casing for engagement of rack and pinion is usually a cast product. However, considering the difficulty of casting, an alternate plan was considered wherein the pinion casing would be substituted by a reducer (2.25”:1.5”).

Since, the smaller section of the reducer was 1.5” (38.1 mm), a ball bearing of appropriate OD was to be selected. A Single-Row O2 Series deep groove ball bearing of bore 17 mm, width 12 mm and OD 40 mm.

As mentioned, a shaft was required to be made for mounting the pinion and bearing. An MS rod of one inch thickness was taken from scrap.

PROCEDURE

Cutting & welding of reducer: The reducer was cut, by a hand hack-saw, on the broader side to provide an opening for the pinion to engage with the rack. This cut reducer was ground in successive iterations for ensuring exact engagement of the rack and pinion and finally welded above the slot in the casing-pipe.

Turning of shaft: The one inch thick rod was turned to achieve a 17 mm thickness for tight fitting the bearing and 10 mm thickness, on the other side, for mounting of pinion.

Modification of reducer

• Cutting with hand hack-saw

Edge preparation

• Hand-grinding

Joining of casing-pipe and reducer

• Arc-welding

11

Mounting: The bearing and the pinion were press-fitted on the shaft and then the shaft was press-fitted into the Hooke’s joint from the side of the bearing. After this assembly, the bearing was press-fitted in the reducer, with help of a sheet-metal sleeve. These press fittings were done manually.

STEERING KNUCKLE

The steering knuckle is mounted between the upper and lower ball joints which connect the steering assembly to the wheels and convert the translational motion of the rack rod to the turning motion of the wheels. The steering arm is connected to the tie rod which goes up to the rack. In our case, the steering knuckle is the end of our model. The hole which connects to the lower ball joint is mounted on the support stand for the demonstration of the model.

MATERIAL

Two MS slabs of 12 mm thickness and 10 mm thickness were used to make the knuckle and steering arm.

PROCEDURE

Shaft preparation

• Turning

Mounting

• Press-fitting

Cutting of plates

•Universal cutting machine

Making of holes

•Drilling

Profiling of Steering arm

•Gas-cutting•Grinding

Joining

•Arc-welding

12

Cutting of plates: The base of the knuckle was made by cutting on the universal cutting machine to get dimensions of 8x4 inches. The two side slabs were cut to dimensions of 3.5 inches and 2 inches in length and 2 inches in width. The 12 mm thick plate was used as the base of the knuckle and the two sides of the knuckle were made from the 10 mm thick plates.

This part involves calculation of caster angle and steering inclination angle. The steering inclination angle used is 12 degrees. The distance between the centers of the two holes is 8.5 inches. The caster angle usually set at about 7 degrees.

Tan(7) x 4.25 = 0.52 inches

Thus, the holes were offset by 0.5 inches on each side from the central axis of the plate.

Making of holes: Two holes of quarter inch each were drilled on the side slabs. The vertical distance between these holes is maintained at 1.8 inches based on the following calculations

Tan(12) = => X=1.8

Profiling of Steering Arm: The steering arm profile was cut as shown in figure alongside. For profiling, the plate was cut by using a carburizing flame during gas-cutting and then ground. Grinding operation in this case was critical and took about 90 minutes of grinding time for each arm.

Joining: The two side plates and the steering arm were joined to the base plate by arc-welding and ground later for better finishing.

SUPPORT STAND AND STEERING ROD

Since the steering assembly thus fabricated has to be setup for demonstration, a support-stand was fabricated. This stand supports the assembly at the casing, knuckle base and steering rod’s bearing.

MATERIAL

Rectangular beams from scrap were obtained. Steering rod and wheel also were obtained from scrap.

4.6”

2.3”

13

PROCEDURE

Support Stand: The critical dimensions for making the support stand include - Height difference between casing and the base of knuckle. Considering an arbitrary height for the knuckle base of 19 inch, the height for the casing support is obtained by adding 4 inches from base of knuckle to steering arm which is at the same level as the casing. Hence, height of 23 inch is required for supporting the casing. Width is decided by taking the distance between the two knuckle-bases which was found to be 46 inch.

Steering Rod: : The steering rod is used connect the steering and the universal cross joint which transmits the rotation from the steering to universal cross joint which in turn rotates the pinion and the drives it. The steering rod is actually not a solid rod but a tube, which is lighter, and has a solid rod section at its end, which enters the Hooke’s Joint.

The rod was assembled in the Hooke’s Joint by grinding the rod thickness to 17 mm, and tightened by a ¼ inch by 2 inch bolt & nut with washer.

On the other side, a solid rod was turned on lathe and welded into the tube for fixing the steering wheel.

TIE-ROD, BALL-JOINT AND HOOKE’S JOINT

For completing the assembly, some components had to be acquired from the old SAE BAJA 2009 on lease. Following section explains their respective functions, and importance in the assembly.

TIE ROD END:

Description: A tie rod is a slender structural unit used as a tie and capable of carrying tensile loads only. Tie rods may be connected at the ends in various ways, but it is desirable that the strength of the connection should be at least equal to the strength of the rod. The ends are threaded and retained by nuts screwed on the ends.

Function: Tie rod is used to join the rack and the steering arm.

Material: High quality alloy steel

Cutting to size

• hand hack-saw• universal

cutting machine

Edge preparation, Profiling

• Grinding

Joining

• Arc welding

14

Specifications of threads:

Unified Screw Threads Fine Series (1)

Unified Screw Threads coarse Series (2)

Size designation = 3/8 Size designation = 3/8

Number of threads per inch = 24 Number of threads per inch = 16

Tensile stress area = 0.0878 sq. inch

Tensile stress area = 0.0775 sq. inch

Minor diameter area = 0.0809 sq .inch

Minor diameter area = 0.0678 sq. inch

BALL JOINT

Description: A ball joint is steel bearing stud and socket enclosed in a steel casing. The bearing stud is tapered and threaded. Ball joints play a critical role in the safe operation of an automobile's steering and suspension.

Function: here the ball joints are used on either side of the rack to join it with the steering arm.

Material: Low carbon steel suitably zinc plated for high strength and good wear resistance.

1

2

15

UNIVERSAL CROSS JOINT:

Description: A universal cross joint or U joint or Hooke's joint is a joint in a rigid rod that allows the rod to 'bend' in any direction, and is commonly used in shafts that transmit rotary motion. It consists of a pair of hinges located close together, oriented at 90° to each other, connected by a cross shaft.

Function: Here the universal cross joint is used as a joint between the steering rod and the shaft which carries the pinion. They are joined by pressing them fit on both sides.

Material: It is generally made of 20Cr steel. Here chromium improves strength, toughness and resistance to corrosion.

Specifications:

1. Inner diameter = 15mm. The shaft is press fit at this side.

2. Inner diameter = 17mm. The steering rod is press fit at this side.

PAINTING AND LUBRICATION

A lubricant is a substance introduced between two moving surfaces to reduce the friction between them, improving efficiency and reducing wear and distribution of heat. Lubricating grease composes of oil base and a thickener. The grease has excellent film strength, improved “anti-washout capabilities”. They possess a higher initial viscosity than oil. Here lubricant is used in the following parts.

1. Pinion is lubricated so that it slides on the rack smoothly.

2. Inside the casing – lubrication is provided inside the casing so that the rack moves to and fro smoothly.

A nipple is attached to the casing to feed lubricant into casing using lubrication gun. There is a small ball that gives a passage for the lubricant when pressure is applied. The ball again comes back to its original position by spring action when the pressure is removed.

1

2

16

PROCEDURE

The nipple has threads with the help of which it fitted. Supposedly, the casing had to be tapped to get the counter-threads. But since the thickness of the casing is not thick enough to make the threads, a hole was drilled and the nipple was brazed onto the casing. And using the lubrication gun the lubricant was fed into the casing.

Painting: The whole assembly is painted to prevent it from getting rusted as it acts as a protective layer from oxidation and dust.

17

REFERENCES

[1] Shigley et. al, Mechanical Engineering Design

[2] B. S. N. Parashar, R. K. Mittal, Elements of Manufacturing Processes

[3] Thomas D. Gillespie, Fundamentals of Vehicle Dynamics

[4] Milliken & Milliken, Race Car Vehicle Dynamics

[5] Carroll Smith, Tune to win

[6] Ghosh & Malik, Manufacturing Science

18

APPENDIX A

The diagram below outlines the important geometry in determining the motions of the wheels in a vehicle that uses Ackerman steering geometry. Ackerman geometry is an interesting problem because it is dynamic. That is to say that we have two components moving together – the left and right steering knuckles, but the relationship between their motion changes as we move them.

Let’s look at the important distances and angles. The two most fundamental distances are the wheel base of the car and the kingpin center to center distance. If we draw two lines representing the wheelbase and the distance from the car’s center line to one of the king pins, we can make a triangle. By design, the line that goes through the centers of the steering arm forms the hypotenuse of this triangle. See below.

Note that the angle with its vertex at A is 90 degrees by design. Note that the line that forms the Ackerman angle with the hypotenuse is parallel with the thrust line. Because of this, we can say that angle B and the Ackerman angle are similar, so if we know one, we know the other. Clearly,

TAN Angle B =

king pin center to center distance / 2 Wheelbase

Wheelbase

King

Pin

Cen

ter

to C

ente

r Di

stan

ce 2

3.25

”

Ackerman Angle

A B

C

19

The problem is that we know the distances and are trying to find angle B. We need the inverse function ARCTAN. Rearranging, we get:

ARCTAN king pin center to center distance / 2 Wheelbase

= Angle B

We can pick distances, turn the crank and find Angle B and by extension, the Ackerman Angle.

According to our design,

Wheel base = 46.5”

king pin to king pin distance = 46.5”. The formula would look as follows:

ARCTAN 46.5” / 2 46.5” ARCTAN (.5) = 26.56º

= Angle B

So, the Ackerman Angle is 26.56 degrees.

20

To find the length of the tie rod, we can decompose the trapezoid ABCD into a rectangle and two triangles.

The distance between the tie rod ends (effective steering system length) is equal to the king pin to king pin center distance minus distance Y on each side. Ackerman arm radius has been chosen to be 5.14”. Well, recall that the SIN of an angle is the ratio between the side opposite the angle and the hypotenuse i.e.

SIN 26.56º = Y/5.14”

Y = 2.3” Now,

LT = DKC – 2*RAA*SIN Ackerman Angle

Where:

LT is distance between the left and right tie rod ends

DKC is the distance between king pins center to center

RAA is the radius of the Ackerman Arm

LT = 46.5” – 2*5.14”*SIN 26.56º

LT = 41.9”

A

D

B

C

A

B 26º56 Y

21

We’ve figured out all of the static values. Now let’s contemplate a turn as diagrammed in red to the right.

According to rack and pinion design the maximum rack travel is 6.3” left to right. As a result of rack travel of 3.15” from centre to one side leads to the inner wheel turning approx. 35º at a turn. Suppose that the Ackerman arm labeled AB steers 35 degrees to the left as shown. Since the car pictured is turning to the left, the right Ackerman arm (CD) needs to steer something less than 35 degrees. To find out how much less let us consider a line drawn diagonally from point D to B. This creates three angles that add together to give the angle of the wheel that pivots at point D. We’ll call the first angle K, the second angle γ (pronounced gamma), and the third angle is of course, the Ackerman angle.

Now we can set to work on determining each. If you think about angle k, we can determine it because for any steer angle, we know the positions of the ends of the diagonal line. If we assigned point A the coordinate of 0,0 then point D would have the coordinates Kingpin Center to Center Distance,0. In the our case specifically point D’s coordinates would be 46.5,0. We can calculate point B’s coordinates with the following formulae:

Point B’s X coordinate = RAA * COS(AA + SAL)

Point B’s Y coordinate = RAA * SIN(AA + SAL)

Where:

RAA is the Ackerman Arm Radius

AA is the Ackerman Angle

B A

C D

35º

Something less than 35º

B A

C D

k

γ

AA

22

SAL is the steering angle of the left wheel. Zero degrees is straight ahead.

Positive values are a left turn, negative values are a right turn. For a 35º left turn,

Point B’s X coordinate = 5.14” * COS(26.56º + 35º)

Point B’s Y coordinate = 5.14” * SIN(26.56º + 35º)

Point B’s X coordinate = 2.45

Point B’s Y coordinate = 4.52

So, the coordinates of Point B at a 35º left turn are 2.45, 4.52. We can project straight to the left of point B and straight up from point A to create a new point called point E. Because we projected straight left and straight up, the angle at E is by definition 90º. Also, because point E falls on segment AD, we can calculate distance DE with the formula:

DE = AD – AE

DE = 46.5” – 4.52”

DE = 41.98”

Now that we know EB and ED, we can find the length of BD because it is a hypotenuse of the triangle formed. Using Pythagorean Theorem:

BD = (EB2 + (DE)2

BD = (2.45”)2 + (41.98”)2

BD = 42.05”

Furthermore, because we know the sides of the triangle we can determine angle k in the following manner:

TAN k = EB/ED

ARCTAN (2.45”/41.98”) = k

K = 3.34º

All we have left is to find angle γ (pronounced gamma). Note that γ is the angle BDC. We know that side DC is the length of the Ackerman arm, which we chose to be 5.14”. We know that side CB is the distance between the tie rod end joints, which we calculated earlier to be 41.9”. Finally, we know the distance BD, which we determined using Pythagorean Theorem to be 42.05”. So we have a triangle and we know the lengths of each of the three sides.

D

A

C D

k

γ

AA

2.45

4.52

E B

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Luckily, there is a somewhat abstract relationship between the sides of non-right triangles called law of cosines. It can be expressed a number of ways, but we will use the permutation shown below.

COS γ = 2AB

A2 + B2 – C2

Per usual, we are trying to get the thing we don’t know by itself, so we’ll need to beat this up a little bit to make it useful. Rearranging gives:

(42.05)2 + (5.14)2 – (41.9)2

2(42.05)(5.14) = COS γ

γ = 84.82º

Now if we add up angle k, γ and the Ackerman angle, we’ll have the tire’s steer angle from the line that connects the two kingpins. To get the steer angle, we have to subtract 90°. The formula is:

Steer Angle = k + γ + Ackerman Angle - 90°

= 3.34º + 84.82° + 26.56° - 90°

= 24.72º

The car is executing a left turn. The left front wheel is steered 35° to the left. The right wheel is tracing a larger arc, and therefore should have a lesser steer angle and hence the right side steering angle should be something less than 35° (which in our case is 24.72º).

The turning circle of a car is the diameter of the circle described by the outside wheels when turning on full lock. There is no hard and fast formula to calculate the turning circle but we can refer to the following formula to get the approximate value.

Turning circle radius = (track/2) + (wheelbase/sin(steer angle))

= (53.5/2) + (46.5/sin24.72) inches

= 3.5m

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APPENDIX B

STEERING KNUCKLE

25

STEERING (ACKERMAN) ARM

26

SHAFT

27

PINION AND BEARING HOUSING

28

RACK

29

PINION

30

RACK AND PINION ASSEMBLY