1Revised 07/10/06

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What is the mission? The mission is to design a Water Rocket Vehicle capable of accurately targeting a specified bullseye.

While promoting Space Propulsion Awareness, the Water Bottle Rocket Competition serves to familiarize students with the basic principles of rocketry, design engineering, manufacturing engineering, and presentation skills.

Students will design and manufacture a water rocket using a 2-Liter bottle as the pressure vessel. The rocket must be capable of accurately targeting a specified bullseye launching from the UTC Rocket Launcher. The design must be supported by technical documentation and presentation outlining all mathematical and scientific principles used. Additionally, each team will develop a patch design, used to symbolically commemorate the objectives of each team. The team’s complete success will not solely be judged on rocket performance, but the combined effort of the team. ...……………………………......GOOD LUCK and Safe Flying !!

*** Remember you will never be a winner unless you try and if you try your best, you have already made it to the bullseye :-) ***

(Refer to Rules & Guidelines and “How to Build Rockets” manual for detailed information.)

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Rules and Guidelines

4

Keys for Safe & Enjoyable Launch Activities:

•Extreme caution should be used at ALL times during launch activity.•The Water Rocket Launcher and Water Rockets are NOT toys; thorough understanding of their function is required prior to conducting ANY launch activities.•Safety Goggles/Glasses shall be worn during all launch activities.•NEVER stand directly in front of launcher at ANY time.•NEVER approach a rocket that is under pressure.•No running, horseplay, etc. around launch area/ during launch activities. Be attentive at ALL times.•Use Only single CLEAR 2L bottles for rocket pressure vessel (Bottles should be in good condition without kinks, cracks or dents).•Always use a CLEAR audible countdown (5!... 4!... 3!... 2!... 1!) before EACH launch.•All Rockets should be thoroughly inspected prior to launch.•Do NOT over-pressurize rockets 80psi MAX.•Use ONLY Large open fields for Launch…..250m or MORE of down range distance is preferred.•Rope-off or clearly mark Launch Area/Field.•Always conduct a trial launch at low air pressure (e.g. 35-40psi) to get an idea of how far your rockets will travel with respect to a

given launch area.

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1. Maximum number of 6 students and a minimum of 4 students per team are allowed.

2. Each team is required to submit a completed technical paper, rocket design, technical drawing and patch design to qualify for the awards. Note: Awards will be presented at the annual FACTRAC banquet.

3. On the day of competition, (but prior to launch) an actual operating rocket with its launching requirements [1. volume of water in milliliters and liters, 2. air pressure in psi (min 30 psi & max 80 psi), 3. Dry Weight in grams, and 4. Nozzel Exit Radius in inches, and 5. Calculated Final Range in meters and feet], corresponding technical drawing, and patch design must be submitted in order to compete in the competition.(Ref. Required note card detailed in Technical Drawing Section, page 12)

Note: 1) At this time each entry must pass a visual inspection and weight requirement

in order to be eligible to compete. Entries that fail inspection will be given ONE opportunity to make modifications to pass inspection, prior to the beginning of the water rocket launching competition.

2) Technical paper must be submitted on _____________________________

4. An overall winner will be judged upon the following criteria detailed on page 17.

5. The objective of the contest is for each team to construct a rocket propelled by water and air which will be launched at a 45 degree angle to hit a target that will bepositioned 61 meters away (distance from HP Rocket Launcher to red bull’s eye; ~200ft), in the target area zone, see Diagram 1a. The target area zone is a 45 degree zone (25 degrees from each side of the target area centerline) if your rocket lands outside of this zone 50 points will be deducted from the accuracy of trajectory score, see Diagram 1b.

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NOTE: Launch accuracy will be scored using the distance and angle from the target the rockets hits in the target area. The scoring equations is as follows:

1875 * 1 + 1 r+30 C+50 Where r = radial distance of rocket from target

C = distance of rocket centerline to launch target area centerline

(see Example below)

Example:Lets imagine five rockets, Rocket A, Rocket B, Rocket D, Rocket E and Rocket F,have launched and this is where they landed ( see Diagram 1b). Rocket D wins First Place for the accuracy of trajectory. Who wins Second Place? Third Place? If rB= rA, Rocket B would win second place for this scenario, as it is closer to the trajectory path. This is because the trajectory path deviation factor “C” . CA is greater than CB, therefore Rocket B wins Second Place in lieu of rB= rA. and Rocket Awins Third Place for accuracy of trajectory , for this scenario. Also Rocket E and Rocket F will have a 50 point deduction for landing outside the target area zone.

7GENERAL AUDIENCE

Launch Command

Launch OperatorLaunch Controller

HP Rocket Launcher

Diagram 1a

** Note: Field markings in FEET.

8

A

B

D

Launch Target Centerline

r

CA

CB

Scoring Example Scenario

Diagram 1b

HP Rocket Launcher

Target Area Zone

E

F

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1. The pressure vessel must be ONE clear 2 liter bottle, see Diagram 2.

2. Water and air pressure will be the sole source of propellant. Before launching rocket, water volume (liters) and air pressure (psi) must be given. NOTE: The air pressure minimum is 30 psi and maximum is 80 psi.

3. Do not use metal, glass, or spikes to construct the rocket. *Use of these materials will automatically disqualify the team from the competition.*

4. On the bottom of the rocket, leave 7.5 cm from the throat of the exit plane clear of any coverings (paint, markings, drawings, etc.), see Diagram 2.

5. Maximum total height of rocket is 76.0 cm, see Diagram 2.

6. Nose-cone tip must have a minimum radius of 1.5 cm, see Diagram 3.

7. No forward swept type of fins are allowed to be used on the rocket.

8. The maximum fin width distance from the bottle is 10.0 cm (or 16.5 cm from center of bottle axis).

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Nose Cone

Fin Fin

Pressure Vessel(Clear 2 Liter Bottle)

Bottle Height(max. 76.0 cm)

Rocket Clear of AnyCoverings (min. 7.5 cm)

Bottle Throat

Diagram 2

Rocket Identification

Throat Exit Plane

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Nose Cone Diagram

Diagram 4

Fin Diagram

Diagram 3

Cone Tip

max 10.0 cm

max 16.5 cm

Min Cone Radius = 1.5 cm

R

12

As part of the competition, the team is required to prepare a scaled drawing depicting the rocket that they have designed and built. 1. The Water Rocket Drawing entry is required to illustrate the actual rocket built by the

team (photographs will NOT be allowed). 2. The size of the engineering paper is required to be the standard 18” X 24”

(Allowing for the 1” margin, the actual drawing is to cover an exposed area of 16” X 22” of the paper). NO MOUNTING, NO FRAMES ALLOWED 3. All dimensions are required to be illustrated on the drawing. 4. The scale and the units are required to be indicated on the drawing. 5. The team’s Water Rocket Drawing is required to show a side and top (or bottom) view. 6. All parts of the rocket are required to be named. 7. A 4” X 5” title card with the following information is required:

Team name and number Team members’ names and discipline Rocket Name (if applicable) Launch Requirements & Specs (Air Pressure (psi), Water Volume (mL

and L) Dry Weight (grams) and Calculated Final Range (m and ft) Picture of Rocket Date

AT COMPETITIONS, THE WATER ROCKET DRAWING WILL BE JUDGED ON: RESEMBLANCE (Between the actual rocket and the drawing) 25 SCALE 25 NAMING/LABELING (Of all the parts used) 25 APPEARANCE/NEATNESS 25 100

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Each entry is to be prepared and submitted by the teams who will be participating in the Water Rocket Design Competition.

Patch designs must be submitted on 8.5”x11” paper.

All entries must contain the team name as well as a detailed explanation of the patch design

All teams participating in the Water Rocket Competition must be prepared to display their patch at the Mini-Design Review.

Patches must be hand-made original work.

Ink pens, pencils, markers or paint may be used.

AT THE COMPETITION, THE PATCH DESIGN WILL BE JUDGED ON:

• ORIGINALITY - Innovativeness of the design. 30

• CREATIVITY - Uniqueness of the information depicted 30

• APPEARANCE - The attractiveness and neatness of the presentation 20

• CONTENT - Design representation of the Team’s name and SECME theme 20

100

What is a “Patch”?

It is a creative display that reflects the dedication and mission of the team.This symbolic picture must comply with the following rules:

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“Here is an Example...”

Explanation of Patch

The propelled rocket represents the school system, supported by the educators and students, following a path towards excellence. The radiant five 4-point stars symbolize the five colleges of Tuskegee University. Where as, the seven 8-point stars represent for the seven business units of United Technologies. The three distinct contrails steaming behind the rocket, symbolize the support offered through Students, Tuskegee University, and UTC. The ring before the rocket depicts the student’s path through the FASTREC program, returning full circle to support the efforts of the program. As we approach the new millennium, the sun over the horizon symbolizes of the induction of the new Water Rocket Design Competition into the FASTREC program. Accuracy, the focus of the contest, is represented by the target created by the outer ring, deep space, and the earth. The border is supported on either side by the chemical symbols respectively, for water and compressed air, which are used to propel the rockets.

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As a part of the Water Rocket Competition, the team is required to write a Technical Report describing the design, construction and operation of the Water Rocket. Reference numbers 1, 2, 3, 4 and 6 are required to be presented together within a maximum of 7 pages. Add pages as appropriate for number 5. Drawings, sketches, and tables may be included in an appendix (optional).

1. COVER PAGE (Required to contain): Title of Technical ReportThe team’s nameNames and disciplines of team membersDate

2. ABSTRACT (One half to one page summary of the Technical Report)

3. INTRODUCTION

4. DESIGN BACKGROUND

5. CALCULATIONS : Table of equations and constantsAssumptionsMass flow rate calculationsDrag calculated assumptionsTrajectory calculations

- location of bottle when all of its water is expelled- final destination point of bottle (Hint: diagram of time vs. distance traveled)

{Calculations will be scored on units, assumptions, accuracy, etc..}

6. CONCLUSIONS/RECOMMENDATIONS

AT THE COMPETITION, THE WATER ROCKET DESIGN TECHNICAL REPORT WILL BE JUDGED ON:

ABSTRACT 10

DESIGN BACKGROUND 10

PAPER STRUCTURE 5

CALCULATIONS 40

CONCLUSION/RECOMMENDATIONS 20

GRAMMAR 15

100 points

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Overall Winner

Best Technical Paper

Accuracy of Trajectory

Best Technical Drawing

Best Patch

Best Design Review Presentation

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Overall Winner:Accuracy of Trajectory 25Technical Paper 35Technical Drawing 15Patch Design 10Design Review Presentation 10Innovative Design 5

100

Points: Best Patch:

Originality (30)Creativity (30)Appearance (20)Content (20)

Accuracy of Trajectory: Target Position (100)

Best Technical Paper:Abstract (10) Design Background (10)Paper Structure (5)Calculations (40) Conclusions (20 )Grammar (15 )

Best Technical Drawing:

Resemblance (25)Scale (25)Name/Labeling (25)Appearance/Neatness (25)

Best Presentation: Effectiveness (40) Creativity (10) Content/Communication (25) Appearance (25)

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Calculations Manual

19

Although you may be anxious to begin building your

rocket, some important decisions need to be made

about launch conditions that can help ensure a more

accurate launch. Remember: the goal of the

competition is to launch the rocket a specific distance. Your

task is to ensure your rocket meets that requirement. The

following calculations will help you determine what choices to

make to predict how far your rocket will go!

These calculations are what is known as an iterative

(repeated) process. First, choose the amount of air

pressure and water to add to the rocket. Then calculate

approximately how far your rocket is predicted to fly. If the resultant

distance is not what you desire, choose another value for air

pressure or water and re-calculate until you reach the answer you

want.

Note: How your rocket really flies will not be exactly what is predicted

here. The actual launch will vary due to real world elements such as wind

changes, drag and differences in your rocket’s physical design that will not be

accounted for now. These factors are left out to simplify calculations. But the

following pages will give you a general starting point for choosing water and

air pressure values and will show you their effect on your rocket’s flight.

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Assume:

Air pressure of the bottle rocket (from 30 to 80 psi) This is the amount of pressure that will be pumped into your rocket at the time of launch.

Choose the mass of water you will use to fuel your rocket. Remember mass= density x Volume. So, for example, if you plan to add 1Liter (volume, about the amount in a sports drink bottle) of water:

Vol= 1Liter= 1000mL= 1000cm3 or 0.001m3 = Density of water= 998 (kg/m3)( a constant)mH2O = x Vol (in kg)

Find:

The Average Mass Flow Rate, , of the water. This is the amount of water (mass) that flows out of the “rocket nozzle”, or throat of the bottle over a period of time ( in a second).

(kg/sec)

Where A= Area of nozzle (m2) = x r2 (r is half the diameter of the 2 liter bottle’s throat)

cd= Discharge Coefficient= 0.98 (a dimensionless constant based on nozzle shape and flow conditions)

= Density of water=998 (kg/m3) ( a constant)P= (1+(Vi/Vf)(Pi)/2. (N/m2) Pi = Initial Air Pressure of bottle (N/m2) Ref: Atmospheric Pressure= 14.7(psi) (or 101,353.56 N/m2)

(a constant)•Final air vol = Vf = 2L

•Initial air vol = Vi = 2L – initial water volume

.

2 PcdAm

m

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Find:

Find the Thrust, ft, of the Rocket. This is the amount of force that pushes the rocket in a forward direction. ( in Newtons) Vmf

t

Find:

Thrust isn’t the only force acting on your rocket. There are also forces acting against the rocket’s motion. The weight (mavex g) of the rocket acts against its attempts to move forward. Drag (fd), (the force of wind acting against the surface of the rocket) also acts on the rocket, but for these calculations, drag will be neglected. (in Newtons)

fd = 0 since the rocket is very small.

Mave= ave. mass of the rocket= [Mempty rocket + Mh2o)/2

g = gravitational acceleration constant= 9.8 m/sec2

gmfffNetForce avedt

Find:

Use your result from to find the Exit Velocity, V, (velocity at the bottle exit), of the water. (m/sec)

A

mV

m

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Scholar’s note: You may be beginning to see how the amount of water you

add affects thrust. The Range equation (equation 8) shows that the higher the

water mass, the longer time the rocket will be propelled, therefore seeming to

increase the Range. So why not just fill the bottle up with water and make it

soar? Well, keep equations 6 and 7 in mind. The mass of the water, has two

functions. It not only increases the time the rocket is propelled, but it also

adds to the force acting against the motion of the rocket (weight), decreasing

acceleration. Too little weight can also be harmful; it can make the rocket

easily affected by the ‘neglected’ elements discussed earlier, like wind

changes. A balance must be achieved.

amave f (force in Newtons: N)

Find:

Find the Acceleration, a (m/s2), of the rocket. Use the equation:

g

VRRange

bottle 2sin2 (in meters)

Where: taVbottle

m

mt OH

2 Time when all of the water is expelled from the Rocket.

angle Rocket is being launched = 45 o

Now find the Range, R, or distance the Rocket will travel, for the water and air pressure conditions you have chosen.

23

There are two remaining factors that will help better determine the actual range of your vehicle. Up until this point we have not taken Drag into consideration for purposes of simplification. It is necessary to account for Drag to predict your rocket’s Final Range (RF). Recall, Drag is the resistance produced on bodies as they move through air. The effects of drag increase with respect to increases in velocity. Determining a Drag Factor To account for the Drag force on your vehicle use the following formula:

D = 1 – (Dc) where Dc : Drag coefficient and D : Drag Factor Select a value for Dc based your nose cone design. See Chart A and calculate D. Note: lower velocities results in lower Dc, (i.e. .15.)

RF = R x D RF is your final distance calculation.

Dc = .21 Dc = .20 Dc = .19 Dc = .23

Now, multiply your Range (R) times the Drag Factor following the equation below:

CHART A

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The effects that the changing water volume and related pressurized air volume will have on your rocket’s performance was accounted for in the initial mass flow rate and acceleration equations.

Why??? Well there are two major reasons to address the changing pressure in the bottle: 1. Actually, the bottle’s internal pressure drops extremely

fast as the water is being expelled and the air volume inside the bottle expands.

Take a moment and think of a balloon that is full and one that is only half full, the air pressure in the half-full balloon is LESS than that in the full balloon. 2. Another important concept to understand is Gas

Compression and Expansion (For our case the gas is Air). Compressed gases within a pressure vessel WILL EXPAND (increase in volume) once the vessel is opened to the atmosphere.

25

In your water rocket problem, the air in your bottle expandsas it pushes the water out. Keep in mind that your bottle hasa fixed volume (2L), therefore, as you INCREASE theVOLUME of WATER, the VOLUME of AIR inside your bottleDECREASES.

Take note:LARGER VOLUMES of GAS will expandMORE than SMALLER VOLUMES of GAS atthe SAME PRESSURE.

Think of the 2L bottles filled with soda you open during apizza party: Even though the internal pressure of the bottlesmay reach as high as 40-50psi, the bottle WILL NOT flywildly out of your hand because the air volume inside,though under significant pressure, is VERY SMALL.

26

So, you’ve calculated the Range, or distance the rocket is

predicted to travel, would the rocket reach the target? Would

the rocket fly too far? Vary the values for water mass and

air pressure. How does the Range Change?

27

How To Build A Water Rocket

28

Newton’s First Law: The Law of Inertia

The Law of Inertia says, “A body in motion remains in motion, a body at rest remains at rest, until acted upon by an outside force.”

Inertia is the tendency to resist any change in motion. It is associated with an object’s mass.

At rest: Forces are balanced. The force of gravity on the rocket balances with the force of the launch pad holding it up.

In Motion: Thrust from the rocket unbalances those forces. The rocket travels upward until it runs out of fuel.

FUNDAMENTAL PRINCIPLES OF ROCKET SCIENCE

HEAVIER rockets have MORE Inertia, because they have MORE mass. MORE Inertia will offerGREATER resistance to a change in direction.Therefore the wind willhave LESS effect on abottle with MORE INERTIA.

A LIGHTER bottle rocket has LESS inertia,because it has LESS mass. LESS inertia means the rocket will have LESS resistance to change in direction. As a result,the wind has a GREATEReffect on the rocket’s path of motion.

WindDirection

Desired Path of Motion

(Trajectory)

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Newton’s second law says: Force = (Mass)(Acceleration)

F = ma

The pressure created inside the rocket is the force (thrust). Mass represents the mass of the rocket and its fuel supply, which in this case is water. Therefore, the mass of the rocket changes during flight. As the fuel is used and expelled, the rocket weighs less and acceleration increases. Thrust continues until the water is completely expelled.

Acceleration

Newton’s Second Law: Force depends upon Mass and Acceleration

F = ma implies that if the forces are the same, then the bigger the mass the smaller the acceleration. The smaller the mass, the larger the acceleration.

Force

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Newton’s third law says, “For every action, there is an equal and opposite re-action.”A rocket takes off when it expels liquid. Action: The rocket pushes liquid outward. Reaction: The liquid exiting the bottle causes the rocket to move in the opposite direction. The Action (Thrust) has to be greater than the weight of the rocket for the Reaction (Liftoff) to happen.

UP

DOWN

(Bottle + Water Mass) x (Bottle velocity)

EQUALS

(Ejected Water Mass) x (Ejected Water velocity)

Essentially, the faster the fluid is ejected, and the more mass that is ejected, the greater the reaction force on the bottle.

Newton’s Third Law: Action and Reaction

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Air Resistance causes Friction which will slow down the rocket.

UP

DOWN

Air Friction (DRAG)

MOTION

MASS EXITING

How to reduce DRAG?

A more pointed nose cone will decrease the air resistance at the front of the rocket, but keep in mind that the minimum nose cone radius is 1/2 inch.

Drag Equivalent to Air Resistance

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The center of mass (CM) is the point at which all of the mass of an object is perfectly balanced. Around this point is where an unstable rocket tumbles.

Balance: Center of Mass and Center of Pressure

The center of pressure (CP) exists only when air is flowing past the moving rocket. Flowing air rubbing and pushing against the rocket can cause it to move around on one of its three axes. It is extremely important that the CP of the rocket is located toward the tail and the CM is located toward the nose.

When the CP and CM are located in the correct place, the rocket will tend to have more stability.

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Brainstorm

The first step in the design of a water bottle rocket is brainstorming. Brainstorming is a problem-solving technique that involves the spontaneous contribution of ideas from all members of the group.

Design Possibilities

The following are illustrations of possible designs for the fins. Any variation of these suggested designs may be used and found to perform better than another when combined with various bottle designs.

DESIGN AND DEVELOPMENT

!Stop! All fins must be at least 4” from the throat exit plane of the bottle (see page 21). This schematic is provided solely to give examples of fin design. We encourage you to be creative.

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Choose best design

Square fins create more stability, but also produce greater drag. Triangular fins introduce less drag, but yield less stability. Taking into consideration the principles of projectile motion, choose the proposed design which best satisfies the objective of the competition.

Design Tips:

Lengthening the rocket makes it more stable by moving the center of mass of the rocket closer to the nose.

Adding fins to the rocket makes it more stable by moving the location where drag forces act on the rocket further to the rear.

Adding mass near the tip of the nose cone makes the rocket more stable by moving the center of mass closer to the nose of the rocket.

Heavier rockets have more inertia; therefore they have more stability. However, remember not too heavy, because the rocket needs to liftoff.

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MATERIALS AND CONSTRUCTION

The following list of materials should NOT be used in any form in the construction of the water rocket. They aredangerous and could cause harm to the operator and those in the presence of the water rocket launch.

Off-limit Materials

Metal

Glass

Spikes and Antennas of any kind.

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Material and Tools Needed

Pressure Vessel (Clear 2-Liter Bottle)

Note: Be certain that your clear, 2-liter bottle is free of scratches, nicks, dents, and discoloration.

AdhesiveFoam mounting tape (approximately 1/16 thick, 2-sided adhesive)Carpet tape (thin 2-sided adhesive)Clear packing tape or Strapping Tape

Use adhesive to bond fins, nose cone, and other allowed materials onto the water rocket

Cutting utensils (Scissors, Hacksaw Blade, Utility Knife, etc.)

Safety First: Children should be supervised at all times while constructing their Water Rockets

For Fin Construction:Balsa and Bass Wood, Cardboard, Plastic, Foam Board, 1/4” to 1/2” thick Styrofoam & Etha Foam, Plastic Plates, and PE (2L) Bottle Material

37

Tip: Using a Low melt glue gun is an excellent way to quickly bond fins. First clearly

mark desired locations on the bottle prior to bonding. Try applying glue to a fin; then apply the fin to one of the marked locations on your bottle. This technique will aid in preventing your pressure vessel (ie. bottle) from deforming due to the ‘initially’ very warm temperature of the glue.

Tip: Using a Low melt glue gun is an excellent way to quickly bond fins. First clearly

mark desired locations on the bottle prior to bonding. Try applying glue to a fin; then apply the fin to one of the marked locations on your bottle. This technique will aid in preventing your pressure vessel (ie. bottle) from deforming due to the ‘initially’ very warm temperature of the glue.

Fin Design & Construction

Determine a fin pattern from your analytic design or trial and error.

Use the recommended materials, however we encourageyou to be creative. Keep in mind not to use the off-limitmaterials.

Cut fins out of the material you choose.

You can use as many fins as you feel are needed.

Attach the fins to the lower section of the rocket using glue, Velcro, tape, or other adhesives.

Tip: It is easier to attach fins to a bottle that is slightly pressurized. You can pressurize the bottle by placing the bottle with its top off in a freezer for 2-3 hours. Next, take it out of the freezer and put the top on very tight, eventually, the air inside warms and the bottle will become slightly pressurized.

BUILDING YOUR WATER ROCKET

38

Typical Fin Patterns

THIS ATTACHED SIDEWILL HAVE THE SAMEPROFILE OF THE SIDE OF 2-LITER BOTTLE

39

Fin Patterns

THIS ATTACHED SIDEWILL HAVE THE SAMEPROFILE OF THE SIDE OF 2-LITER BOTTLE

40

More Fin Patterns

THIS ATTACHED SIDEWILL HAVE THE SAMEPROFILE OF THE SIDE OF 2-LITER BOTTLE

41

Nose Cone Design & Construction:

Determine what material you want to use.

Pattern the nose cone and cut it out.

Attach the nose cone to the top of the rocket by using some recommended adhesives.

Note: Remember use only the material recommended and maintain a nose radius

of 0.5 inch or greater.

Tip: Add ballast (weight) to nose cone (e.g. Styrofoam-peanuts, shredded paper, etc.) to shift the water rocket’s center of mass forward and increase its flight stability. Smaller amounts of more dense materials such as clay, sand, water, etc. may also be used as ballast. Remember not to use the Off-Limit materials.

Nose Cone Design & Construction:

Determine what material you want to use.

Pattern the nose cone and cut it out.

Attach the nose cone to the top of the rocket by using some recommended adhesives.

Note: Remember use only the material recommended and maintain a nose radius

of 0.5 inch or greater.

Tip: Add ballast (weight) to nose cone (e.g. Styrofoam-peanuts, shredded paper, etc.) to shift the water rocket’s center of mass forward and increase its flight stability. Smaller amounts of more dense materials such as clay, sand, water, etc. may also be used as ballast. Remember not to use the Off-Limit materials.

42

Preferred Nose Cone Construction- Water Rocket Assembly Method

Step 1: Cut the bottomoff of a 2L Bottle(discard bottom). Step 2: Carefully align

top portion of bottle on the 2L bottle to be used for the pressure vessel.

Step 4: Tape/secure the joint between the nose cone stage and the pressure vessel.

Step 3: Rotate and observe your water rocket from several angles to ensure good alignment.

Tips: - Remember to add ballast to your nose cone stage. - The pressure vessel should be in good condition free of scratches and dents)

43

Option 2: Ballast can be added before or after you permanently affix the nose cone to the pressure vessel.

Option 1: A) The neck can be cut off of the top of the nose cone as shown in Step 4; this will slightly improve the aerodynamics of the rocket. B) The resulting hole can simply be covered with tape. (Use a hack saw blade for cutting through the thicker materialat the neck of bottle. Utility and other knives are NOT recommended for thisprocess).

BEFORE AFTER

44

Option 3: The length of the Rocket can be increased by adding another bottle between the nose cone stage and the pressure vessel. Remember to stay within the dimensional limits for the competition.

Note: Taller rockets will notnecessarily perform better than shorter ones. Try to keep your construction/assembly process as simple as possible.

45

Alternative Example of Nose Cone Construction

Step 1: Cut a Circleout of thick stock paperor thin poster material(Using 16” or larger diameter).

Step 2: Cut a line along the radiusas shown.

Step 3: Rotate the paper into a cone. Next Tape or Gluethe seam to maintain the cone’sshape. You can adjust angle of the cone with more rotation. (Keep mind that the base of yourcone needs to be large enoughto fit around the top of thepressure vessel).

Step 4: If needed trim the base of cone as required so that it has a uniform fit with the diameter of a 2L bottle.

Uniform Fit All-Around Here

46

Step 8: Secure the resulting nose cone to the pressurevessel using an adhesive like tape, glue, velcro etc...

Step 6: Uniformly trim topof paper nose cone to accept a craft foam or styrofoam ball or cone.

Be certain to use some form of ballast (weight) to shift your rocket’s center of mass forward.

Step 7: Add the foam ball or cone to create a 0.5” or larger nose cone radius.

47

More on Water Rocket Construction:

A) For lengthened rockets (Option 3 Page 20) A piece of 1/2” PVC Pipe can be used to align the nose cone to a second bottle prior to assembly with the main pressure vessel bottle.

B) Join the bottles together on the PVC shaft and tape the joint between bottles securely. (Make certain tape lays flat on the bottle’s surface).

C) Now, remove the PVC shaft and join upper nose cone stage to the pressure vessel. Carefully align the the stages.

(Note: you will NOT be able to use the PVC shaft toalign the nose cone and attached bottle to the pressure vessel).

Other Tips

A B C

Pressure Vessel

48

FINS: Whether your fins are wide or thin the primary ‘assembly’ objectives/considerations should be:

1) Make certain fins are aligned with center axis of rocket.

2) Be sure fins are well affixed to bottle to prevent separation or deflection/movement during flight.

3) Wider fins (1/4”-1/2” thick) provide a larger attachment/ contact surface. They can be securely attached using tape only and are useful for quick assembly & especially when working with young children due to ease of assembly.

4)Thinner fins (3/16” or less) are excellent for reducing the effects of drag, however, more effort is usually involved with securely attaching them to your water rocket. Thin fins must be very stiff once mounted to prevent movement during flight.

49

5) A minimum of three fins are recommend for stable flight (4 fins are a good choice as well)

6) All fins should be spaced equally apart regardless of the number (e.g. 3 fins-120o apart, 4 fins-90o apart, and so on).

Note: Aligned fins are recommended, particularly when competing. Tilting fins will cause rockets to spin. This action may slightly increase flight stability but will likely make it more difficult to ‘calculate’ how far the rocket will travel. In case fins are tilted to cause ‘spin’:• They must ALL be tilted in the same direction.• They should only be tilted slightly (e.g. 2o to 10o).• The fins should be equally spaced.• It is strongly suggested that you try the aligned fin approach first!!!

50

Examples Section

51

M a s s F l o w R a t eM a s s F l o w r a t e i s a m e a s u r e o f t h e a m o u n t o f m a s s ( F l u i d ) p a s s i n g t h r o u g ha g i v e n a r e a w i t h r e s p e c t t o t i m e . S o m e e v e r y d a y e x a m p l e s o f m a s s f l o wr a t e a r e w a t e r t r a v e l i n g t h r o u g h a f i r e m a n ’ s h o s e , s o d a f l o w i n g f r o m af o u n t a i n i n t o a c u p , a n d p r o p e l l a n t b e i n g r a p i d l y e x p e l l e d f r o m a r o c k e t ’ se n g i n e s . T h e m a s s f l o w r a t e o f a g i v e n f l u i d c a n b e d e t e r m i n e d w i t h t h ef o l l o w i n g e q u a t i o n :

E x a m p l e 1A m o t h e r n e e d s t o f i l l a b a t h t u b h a l f f u l l s o t h a t h e r y o u n g d a u g h t e r c a n t a k e a b a t h . T h et u b h a s a t o t a l c a p a c i t y o f 8 0 l i t e r s . I f w a t e r f l o w s t h r o u g h t h e n o z z l e i n t o t h e t u b a t f l o wr a t e o f . 2 0 k g / s e c h o w l o n g w i l l i t t a k e t o f i l l t h e t u b h a l f w a y ?

S t e p 1 D e t e r m i n e h a l f t h e t u b ’ s v o l u m e c a p a c i t y . T o t a l c a p a c i t y = 8 0 L ,s o t h e h a l f c a p a c i t y = 4 0 L .

S t e p 2 N e x t w e m u s t c a l c u l a t e m a s s f o r 4 0 L o f w a t e r . C o n v e r t t h e w a t e r v o l u m et o m a s s b y m u l t i p l y i n g t h e v o l u m e o f w a t e r r e q u i r e d b y t h e d e n s i t y o fw a t e r . ( i s p r o n o u n c e d ‘ r h o ’ ) .

= 9 9 8 k g / m 3

W e k n o w t h a t 1 L = . 0 0 1 m 3

t h e r e f o r e , 4 0 L = . 0 4 m 3

n o w m u l t i p l y , . 0 4 m 3 x 9 9 8 k g / m 3 = 3 9 . 9 2 k g

S t e p 3 N o w t h a t w e k n o w t h e m a s s o f w a t e r r e q u i r e d w e c a n d e t e r m i n e t h ea m o u n t o f t i m e ( t ) r e q u i r e d t o f i l l t h e t u b h a l f f u l l b a s e d o n t h e d e f i n e dm a s s f l o w r a t e o f . 2 0 k g / s e c , t h a t i s f o r e v e r y s e c o n d t h a t p a s s e s . 2 0 k g o fw a t e r w i l l f l o w i n t o t h e t u b .

t = sec)/(

)(2

kgm

kgm OH

t = sec)/(20.

)(92.39

kg

kg

t = 1 9 9 . 6 s e c o r 3 m i n 1 9 . 6 s e c

C h a l l e n g e : I f t h e m a s s f l o w r a t e f o r t h e a b o v e e x a m p l e e q u a l s . 2 0 k g / s e c ,

w h a t i s t h e v o l u m e f l o w r a t e e q u a l t o ?

.

2 PcdAm

52

Example 2A gardener must water his garden daily due to a severe drought. It is important that his smallcrop of vegetables get at least 200L of water each day. He uses a nozzle attached to a hosethat supplies water at a pressure of 25psi. Considering the nozzle has an exit diameter of 2cm,determine the mass flow rate.

Comprehension:It is critical to understand exactly what is occurring during this process. While the water is beingsupplied at a pressure (Ps) of 25 psi, it is being expelled into a pressurized environment ‘OurAtmosphere’. While atmospheric pressure varies, it is safe to assume that atmospheric pressurePatm equals 14.7 psi for this sea level application. This pressure will offer resistance to the waterbeing ejected from the nozzle and therefore must be accounted for. Hence, the pressuredifference or P (pronounced ‘delta’ P) is derived by subtracting the atmospheric pressure Patm

or (14.7psi) from the supply pressure Ps (25psi). We will useP for the mass flow ratecalculation.

Likewise, it is necessary to calculate P for your rocket’s mass flow rate equation based on its initial ‘vessel’ or internal pressure and Patm

Step 1 Given supply pressure (Ps) and atmospheric pressure (Patm) calculate P:

Ps = 25psiPatm = 14.7psi

P = Ps - Patm P = 10.3psi

For this metric calculation, psi (pounds square inch) must be converted toN/m2 (Newton per square meter) so multiply 10.3 psi by 6.8948 x 103 or 6894.8,

P = 71016.4 N/m2

Step 2 We know the density of water H2O is 998 kg/m3 at room temperature

Step 3 Calculate the ‘effective flow area’ for your nozzle. First, determine the area for agardener’s nozzle having a 2cm exit diameter. Multiply the result by .98, thedischarge coefficient or cd.

About Discharge Coefficients: Discharge Coefficients are used to account for flow lossescaused by non-uniform flow paths. Since a gardener’s nozzle converges down to 2cmfrom 4cm it is reasonable to assume that the flow rate will be reduced by a dimensionlessfactor of .98 or 2 percent due to the change in flow area.

You will be required to apply a discharge coefficient during the mass flow rate calculation for your rocket, due to the converging nozzle of your 2L bottle.

53

Example 2 Continued…………….Area for a circle (A) = r2

where = 3.14 and d=2cm or .02m r = d/2 = 1cm

and 1cm = .01m A = x .(01m)2

A = 3.14 x .1000m2

A = .000314m2

thus, Acd = .000314m2 x .98 Acd = .000308m2

Step 4 Now calculate the mass flow rate given:

P = .01 N/m2

H2O = 998 kg/m3

Acd = .000308m2

m = .000308m2 x 234.710169982

m

Nx

m

kgx

Keep in mind that 1N = 1 2sec

mkg

That is one Newton equals the acceleration of 1 m/sec 2 to a one kilogram mass.

m = .000308m2 x 2

2

3sec4.710169982m

mkg

xm

kgx

Now pay close attention to what happens to the units……

m = .000308m2 x 24

27

sec1017.14

m

kgx

m = .000308m2 x 11905.8 sec2 m

kg

m = .000308m2 x 11905.8 sec2 m

kg

m = 3.67 sec

kg

.

2 PcdAm

54

T h r u s tJ e t a n d r o c k e t e n g in e s c r e a te th r u s t b y a c c e le r a t in g p r o p e l la n ts ( u s u a ll y h o t g a s e s ) to h ig hs p e e d s . O th e r o b je c ts c a n c r e a te f o r c e s in s im ila r w a y s , th o u g h . A s N e w to n ’s T h ir d L a w s ta te s ,a l l a c t io n f o r c e s c r e a te a n e q u a l a n d o p p o s ite r e a c t io n f o r c e . T h u s a n y o b je c t t h a t c a u s e s m a s sto a c c e le r a te in o n e d ir e c t io n w i l l e x p e r ie n c e a ‘ th r u s t ’ f o r c e ( f t) in th e o p p o s it e d ir e c t io n . T h ea m o u n t o f f o r c e is d e s c r ib e d b y t h e t h r u s t e q u a t io n :

f t =

m V

w h e r e

m is th e m a s s f lo w r a te ( in k i lo g r a m s /s e c o n d ) o f th e p r o p e l la n ts o u t o f th e o b je c t a n d V isth e v e lo c it y ( in m e te r s /s e c o n d ) th a t i t e x its . N o te t h a t i f th e e x it in g p r o p e l la n ts a r e h ig h - p r e s s u r ef lu id s e x it in g in to t h e a ir , t h r u s t w il l b e s o m e w h a t h ig h e r . M o s t o f th e t im e , th o u g h , w e c a na s s u m e th a t a je t o f p r o p e l la n ts e x its a t a tm o s p h e r ic p r e s s u r e a n d u s e th e a b o v e e q u a t io n .

E x a m p le 3

F ir e f ig h te r s o f te n n e e d to s p r a y la r g e a m o u n ts o f w a te r to g r e a t h e ig h ts . T o d o s o , th e yu s e h ig h - p o w e r p u m p s a n d h e a v y - d u ty h o s e s th a t a c c e le r a te th e w a te r to h ig h s p e e d s .T h is s p r a y in g c r e a te s a r e a c t io n fo r c e o n th e h o s e th a t c o u ld c a u s e i t t o m o v eb a c k w a r d s v io le n t ly . T w o o r m o r e f ir e f ig h te r s th u s o f te n h o ld a f ir e h o s e s te a d y toc o u n te r a c t th is th r u s t .

C o n s id e r o n e s u c h f ir e h o s e a t ta c h e d to a t r u c k th a t p u m p s w a te r a t a m a s s f lo w r a te o f8 0 k g /s . T h e d ia m e te r ( d ) o f t h e n o z z le is . 1 m e te rs . F in d th e th r u s t f o r c e o n th e h o s ec r e a te d b y th e s p r a y in g o f w a te r ( d e n s it y = 9 9 8 k g /m 3 ) . Y o u m a y a s s u m e th a t t h e w a te re x it s th e h o s e a t a tm o s p h e r ic p r e s s u r e .

S o lu t io n

T h e e q u a t io n fo r t h r u s t is :

f t =

m V

w h e r e

m is t h e m a s s f lo w r a te a n d V is t h e v e lo c it y o f t h e w a te r e x it in g th e n o z z le .T h o u g h w e a r e o n ly g iv e n th e v a lu e o f m d o t , w e c a n f in d V f r o m th e e q u a t io n :

V = )( A

m

w h e r e is t h e d e n s it y o f t h e w a te r . T o f in d V w e m u s t f i r s t d e te rm in e th e e x it a r e a o f t h en o z z le . S in c e w e k n o w th a t a r e a = R 2 w h e r e R is h a lf t h e n o z z le d ia m e te r , w e h a v e :

A = r 2

A = ( d / 2 ) 2

A = ( .1 m / 2 ) 2

A = .0 0 8 m 2

55

Example 3 Continued……………….

Now we can find the exit velocity:

V = )( A

m

V = 80 kg/s / (998 kg/m3 .008 m2)

V = 10.0 m/sec

Now we know mass flow rate and velocity so we can find thrust:

ft =

m V ft = 80 kg/s 10.0 m/sec

ft = 801 N

Thus the firefighters have to put almost 200 lbs of force on the hose to keep it fromaccelerating backwards!

Now take a moment to apply the above theory to your water rocket:Gravity acts on the mass of your rocket and keeps at rest on the launcher,at least until another force acts on the rocket. In order for the rocket tomove, the other force acting on it MUST be GREATER than earth’sgravitational pull. In Example 3 think of the force the firemen apply to thehose as gravity or a ‘holding force’. Unlike the firemen’s success withkeeping the hose in position, the thrust force produced by the water rapidlyexiting from your bottle will overcome the effects of gravity on the totalmass of the rocket…………at least temporarily. Since the thrust force (ft)is applied only for a short time, gravity will eventually win causing therocket to return to earth.

56

Force & Acceleration

Newton’s First Law of motion tells us that an object will only accelerate when aforce is applied to it. Be careful, though! Often times, forces will cancel eachother by acting in opposite directions. For example, your weight is a force that ispulling you toward the center of the earth. The chair you are sitting in, however,is exerting an upward support force that is exactly equal to your weight. Thusthe net force is zero and you do not experience an acceleration.

When the net force on an object is not zero—i.e. when unbalanced forces exist—the object will accelerate. Newton’s Second Law states this acceleration by thefollowing equation:

F = ma

Solving for acceleration, a = F / m

Where F is the net force in Newtons, m is the mass in kilograms, and a is theacceleration in meters/second2. Note that a and F will always act in the samedirection.

Example 4

Solution:

The two forces acting on the shuttle are its weight and the thrust of the engines.The weight force W is simply the mass times the acceleration of gravity:

W = -mg

W = -(2,700,000 kg)(9.8 m/s2)

W = -26,460,000 N (downwards)

Rocket engines create thrust through the principle ofreaction forces. In the previous example, the firemenapplied a reaction force equal to the thrust forcegenerated by the rapid mass flow rate of water beingexpelled. In this case, rocket engines acceleratepropellants (usually hot gases) downwards to create anupward reaction force.

Consider the Space Shuttle, which weighs approximately2,700,000 kg as it sits on the launch pad. Once the mainengines and solid-rocket boosters have started andreached full power, they produce a total of 34,400,000N ofthrust. The shuttle does not move upward, however, untilexplosive bolts release the boosters. Recalling thatacceleration is a result of the net force acting on anobject, calculate the instantaneous upward acceleration ofthe shuttle when the bolts release.

57

Example 4 Continued……………

Again, Fnet = Ft + W

Fnet = 34,400,000 N + (-26,460,000 N)

therefore, Fnet = 7,940,000 N

From the Second Law, we can thus find the initial acceleration:

Fnet = F

F = ma

a = F / m

a = 7,940,000 N / 2,700,000 kg

a = 2.94 m/s2

Note that the mass of the shuttle is actually constantly changing: propellants are beingexpelled from the engines at high speed. Thus the acceleration will continue to increasedramatically as the shuttle lifts off.

58

A c c e l e r a t i o n a n d V e l o c i t y

W e h a v e j u s t s e e n h o w N e w t o n ’ s l a w s h e l p d e s c r i b e t h e r e l a t i o n s h i p b e t w e e na n o b j e c t ’ s a c c e l e r a t i o n a n d t h e f o r c e s a c t i n g o n i t . T h i s i s i m p o r t a n t b e c a u s ew e h o p e t o p r e d i c t t h e m o t i o n o f o u r r o c k e t a n d f i n d i t s r a n g e . B u t h o w d o e sk n o w i n g t h e a c c e l e r a t i o n h e l p u s ? R e m e m b e r t h a t a c c e l e r a t i o n i s s i m p l y h o wf a s t a n o b j e c t ’ s v e l o c i t y i s c h a n g i n g a t t h a t m o m e n t . I f w e c a n a s s u m e t h a ta c c e l e r a t i o n i s c o n s t a n t d u r i n g a t i m e p e r i o d t , t h e n w e c a n f i n d t h e c h a n g e i na n o b j e c t ’ s v e l o c i t y ( v - v o ) :

v - v o = a t ( 3 )

c h a n g e i n v e l o c i t y = a c c e l e r a t i o n t i m e

o r w e c a n r e w r i t e t h e e q u a t i o n :

v = a t + v o ( 4 )

f i n a l v e l o c i t y = a c c e l e r a t i o n t i m e + i n i t i a l v e l o c i t y

T h i s s h o w s u s t h a t w e o n l y n e e d t o k n o w t h e i n i t i a l v e l o c i t y a n d a c c e l e r a t i o n t of i n d t h e f i n a l v e l o c i t y a f t e r t i m e t .

E x a m p l e 5

S o l u t i o n :

T h e t i m e t c a n b e f o u n d b y r e a r r a n g i n g e q u a t i o n ( 4 ) a n d s u b s t i t u t i n g v = 0( b e c a u s e t h e c a r w i l l b e s t o p p e d a f t e r t i m e ( t ) . T h e a c c e l e r a t i o n a = - 6 m / s 2 i sn e g a t i v e b e c a u s e i t i s i n t h e d i r e c t i o n o p p o s i t e o f t h e c a r ’ s v e l o c i t y :

v = a t + v o

t = ( v - v o ) / a

t = - 6 2 f t / s / - 6 f t / s

t = 1 0 . 3 s e c o n d s

A s p o r t s c a r i s t r a v e l i n g f o r w a r d a t 6 2 f t / s w h e nt h e d r i v e r l i g h t l y a p p l i e s t h e b r a k e s . T h e b r a k e sc a u s e a c o n s t a n t d e c e l e r a t i o n o f 6 f t / s 2 . H o wm u c h t i m e w i l l i t t a k e t h e c a r t o c o m e t o as t o p ?

59

R a n g eR a n g e i s t h e d i s t a n c e a n o b j e c t w i l l t r a v e l d e p e n d i n g o n i t s v e l o c i t y a n d a n g l e o ft r a j e c t o r y ( ) . T h u s t w o o b j e c t s h a v i n g t h e s a m e m a s s w i l l t r a v e l t h e s a m e d i s t a n c eu n l e s s a c t e d u p o n b y o u t s i d e f o r c e s s u c h a i r r e s i s t a n c e ( d r a g ) ; t h i s f a c t e x p l a i n s w h y i ti s d i f f i c u l t t o t o s s a b a l l o o n f i l l e d w i t h a i r , b u t r a t h e r e a s y t o t o s s a b a l l o o n p a r t i a l l y f i l l e dw i t h w a t e r . K e e p i n m i n d t h a t m o r e f o r c e i s r e q u i r e d t o a c c e l e r a t e a h e a v i e r m a s s t o t h es a m e v e l o c i t y a s t h a t o f a l i g h t e r m a s s .

T h e d r y m a s s o f y o u r r o c k e t i s c r i t i c a l f o r s t a b i l i t y a n d f o r o v e r c o m i n g d r a g , b u te x c l u d i n g d r a g r o c k e t s h a v i n g t h e s a m e v e l o c i t y ( s p e e d ) w o u l d t r a v e l t h e s a m ed i s t a n c e . A i r r e s i s t a n c e w i l l h a v e a g r e a t e r e f f e c t o n t h e r a n g e o f l i g h t e r m a s s e st h a n l a r g e r m a s s e s m o v i n g a t t h e s a m e v e l o c i t y .

E x a m p l e 6A w o m e n p l a y i n g c e n t e r f i e l d m u s t q u i c k l y t h r o w a s o f t b a l l t o s e c o n d b a s e d u r i n g ag a m e t o p r e v e n t a p l a y e r o n t h e o p p o s i n g t e a m f r o m a d v a n c i n g f r o m f i r s t b a s e . I f s h er e l e a s e s t h e b a l l f r o m h e r h a n d a t a n a n g l e o f 3 0 o , a t w h a t v e l o c i t y m u s t t h e b a l l b et h r o w n i n o r d e r t o r e a c h s e c o n d b a s e w h i c h i s 5 0 m a w a y ?

N o t e : F o r s i m p l i c i t y w e w i l l i g n o r e t h e d r a g f o r c e o f ‘ a i r ’ i n t h i s e x a m p l e , h o w e v e r , d r a gh a s a s i g n i f i c a n t e f f e c t w i t h m a n y t r a j e c t o r y a p p l i c a t i o n s , i n c l u d i n g t h e t r a j e c t o r yc a l c u l a t i o n f o r y o u r r o c k e t .

U s i n g t h e R a n g e E q u a t i o n

R = g

xV sin2

w h e r e , R : R a n g e = 5 0 mV : V e l o c i t y = ? ( m / s e c )T : R e l e a s e o r ‘ T r a j e c t o r y ’ A n g l e = 3 0 o

G : G r a v i t a t i o n a l A c c e l e r a t i o n = 9 . 8 1 m / s e c 2

D e t e r m i n e V e l o c i t y :

F i r s t w e m u s t r e a r r a n g e t h e r a n g e e q u a t i o n a s s h o w n :

V 2 = 2sin

gxR

V = 2sin

gxR

60

Example 6 Continued……………

V = ox

mxm

302sinsec

81.950 2

V = 8660.

sec5.490

2

2m

V = 2

2

sec4.566

m

therefore, V = sec

8.23m

Wow that’s neat but I relate better to miles per hour (mph)……………..

Okay, to convert meters per second to miles per hour use the following relationship:

1m/sec = 2.2369 mph

So we multiply the result by 2.2369

23.8m/sec x 2.2369 = 53.2 mph!

Example 7

A high school quarterback passes a football at velocity of 20m/sec to a receiverrunning down field directly in front of him. If the quarterback releases the football at anangle () of 45o, what distance must the receiver reach to catch the ball? (Assume thatthe receiver will catch the ball at the same height that it was thrown from.)

Again:

R = g

xV 2sin2

61

Example 7 Continued…………………

quickly, V= 30m/sec= 45o

g = 9.81 m/sec2

R =

2

2

sec81.9

2sinsec

20

m

xm

R =

2

2

2

sec81.9

90sinsec

400

m

xm o

Considering sin90o = 1, R =

2

2

2

sec81.9

sec400

m

m

R = 40.8m

Great, but I want to know the equivalent of 40.8m in feet (ft).

Well we know 1m = 3.2808 ft

so simply multiply 40m x 3.2808

We have R= 133.8ft

Congratulations on reading through this brief Appendix. These examplesare intended to help you better understand how many of the physicalprinciples of Rocketry relate to everyday applications. We encourage youto create and work through problems of your own to further stimulate yourunderstanding of the concepts of Mass Flow, Thrust, The Laws of Motion,and Trajectory Calculation. More in depth equations and examples willfollow in future editions of this manual. Good Luck Rockeeters as you aimfor the Stars!

Note: A useful conversion table is provided on the following page.

62

Conversion Table

Multiply By To obtaincentimeter (cm) 3.2808 x 10-2 feet

3.9370 x 10-1 inches1.0000 x 10-2 meters

cubic centimeter (cm3) 6.1024 x 10-2 cubic inches1.0000 x 10-6 cubic meters

cubic foot (ft3) 2.8317 x 104 cubic centimeters1.7280 x 103 cubic inches2.8317 x 10-2 cubic meters

cubic inch (in3) 5.7870 x 10-4 cubic feet1.6387 x 10-5 cubic meters

cubic meter (m3) 1.0000 x 106 cubic centimeters3.5315 x 10 cubic feet6.1024 x 104 cubic inches

foot (ft) 3.0480 x 10 centimeters3.0480 x 10-1 meters

foot/second (fps) 1.0973 kilometers/hour3.0480 x 10-1 meter/second6.8182 x 10-1 miles/hour

inch (in) 2.5400 centimeters2.54 x 10-2 meters

kilogram (kg) 1.0000 x 103 grams3.5274 x 10 ounces2.2046 pounds9.8067 newtons

kilogram/square meter(kg/m2) 9.8067 newtons/square meterliter (l) 3.5315 x 10-2 cubic feet

2.6417 x 10-1 gallons (U.S. Liquid)1.0000 x 10-3 cubic meters

meter (m) 1.0000 x 102 centimeters3.2808 feet

meter/second (m/sec) 3.2808 feet/second3.6000 kilometers/hour2.2369 miles/hour

mile/hour 1.4667 feet/second1.6093 kilometers/hour4.4704 x 10-1 meters/second

newton (N) 1.0197 x 102 grams1.0197 x 10-1 kilograms2.2481 x 10-1 pounds

Please Note: The above Conversion Table is provided as an aid. Use of the all of conversionfactors is not required for the trajectory calculations. Be careful! Pay attention to units andexponents. Make sure you use only those conversions which are needed for your calculations.

63

newton/square meternewton/square meter(pascal (Pa)) (N/m2) 1.0197 x 10-1 kilograms/square meter

2.0885 x 10-2 pounds/square foot1.4504 x 10-4 pounds/square inch

ounce (oz) 2.8349 x 10 grams2.8349 x 10-2 kilograms6.2500 x 10-2 pounds

pound (mass) (lb) 4.5359 x 102 grams4.5359 x 10-1 kilograms1.6000 x 10 ounces

pound (force) (lbf) 4.4482 newtons4.4482 kilonewtons

pound/square inch (psi) 7.0307 x 102 kilograms/square meter6.8948 x 103 newtons/square meter1.4400 x 102 pounds/square foot

square foot (ft2) 1.4400 x 102 square inches9.2903 x 10-2 square meters

Please Note: The above Conversion Table is provided as an aid. Use of the all of conversionfactors is not required for the trajectory calculations. Be careful! Pay attention to units andexponents. Make sure you use only those conversions which are needed for your calculations.

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Construction Help

65

Building Fins From 2-Liter Bottles

1. Cut Top and Bottom Off 2. Flatten and Cut

3. Reverse the Fold and Recrease

Note: The method of design and construction shown here is only an example. Use your imagination to create new designs using the recommended materials.

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An Example Only:Building Fins From 2-Liter Bottles

5. Trim to Desire Sweep, Add Clear PackingTape Over Trailing Edge.

4. Add Double Side Tape

Thin Carpet Tape(Trailing Edge)

Thick MountingTape Center Spar

Note: Adding clear packing tape keeps the leading edge from curlingup and mounting tape add strength and stiffness to the fin.

Tip: Add a smooth fillet of glue around the base of each fin.

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Nose Cone

Fin Fin

Pressure Vessel(Clear 2 Liter Bottle)

Bottle Height(max. 30 inches)

Rocket Clear of AnyCoverings (min. 3 inches)

Fins Start(min. 4 inches)Bottle Throat

Diagram 1

Rocket IdentificationBallast Added to theNose Cone (e.g. Styrofoam-peanuts, shredded paper, etc.)

Throat Exit Plane

Min Cone Radius = 0.5 inches

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Nose Cone Diagram

Diagram 3

Fin Diagram

Diagram 2

Cone Tip

max 10.2 cm

Min Cone Radius = 0.5 inches

R Note: Make certain to construct the tip of the nose coneper the minimumcone radius(0.5 inches) for safeoperation.

max 16.5 cm