Fill This Form To Receive Instant Help

Help in Homework
trustpilot ratings
google ratings


Homework answers / question archive / UASE 501 Worksheet 7

UASE 501 Worksheet 7

Mechanical Engineering

UASE 501 Worksheet 7.2

Worksheet: UAS Launch and Recovery

UAS 501: Introduction to Unmanned Aircraft Design

Embry-Riddle Aeronautical University

  1. Introduction

 

This worksheet reviews and clarify some of the equations as a supplement to Chapter 11 of Gundlach’s textbook Designing Unmanned Aircraft Systems. First, an overview of general equations used within the text are presented. The worksheet then addresses key equations for several launch and recovery approaches discussed within the text.

 

Exercises are distributed throughout the worksheet provide an opportunity to apply the calculations to realistic problems.

 

This worksheet is not a replacement for the textbook. The textbook provides alternative approaches, equations, and key insights to help guide your analysis and better understand the benefits and constraints of each approach.

 

    1. Note on Units

 

The equations presented can be applied to both SI and English units. Please remember that weight is a force. In British Imperial units, a pound is a unit of force (represented as lb or lbf). In SI units, newtons are the unit for force. With respect to energy, the SI unit is joules, and the British Imperial unit is foot-pound force (equal to 1.3558 J).

 

Within this worksheet, the units specified are give such that the SI unit is on the left and the British Imperial unit is on the right.

 

  1. General Equations

 

The following are some general equations that have broad applicability across different launch and recovery systems directly or with some adaptation.

 

    1. Impulse-Momentum Equation:

 

The Impulse-Momentum Equation defines the relationship between an impulse, i.e. change in force over time, and change momentum. For constant force over the period of time, the relationship is represented as:

 

F⋅Δt=WTOg|ΔV|

 

 

 

Where,

  • F
     – force (N
     or lbf
    )
  • Δt
     – duration of force (s
    )
  • WTO
     – takeoff weight (N
     or lbf
    )
  • g
     – gravity constant (9.8 m/s2
     or 32.17 ft/s2
    )
  • ΔV
     – change in velocity (m/s
     or ft/s
    )

 

 

This equation is valuable because it defines a relationship between the force associated with the launch or recovery event and parameters of that event including the weight of the aircraft at the initiation of the event and the change in velocity magnitude over the duration of the event.

 

Note: the equation above includes WTO

 for takeoff weight, but this relationship can be generalized to recovery as well using recovery weight instead.

 

 

 

Exercise:

 

  1. A winch released UAS must be launched with a minimum airspeed of 3 m/s. The vehicle is launched at 5 meters and the vehicle must attain its launch velocity and climb out above 0.5 meters. The target launch velocity is 15 m/s.
    • What is the maximum time permitted from release to attaining the vehicle’s launch velocity before reaching the minimum altitude threshold?
    • Using the impulse-momentum equation, determine the minimum line tension to attain the launch velocity before reaching the minimum altitude threshold?

 

 

 

    1. Stroke Length Acceleration

 

The chapter frequently utilizes the following equation to specify the length of a motion or stroke such as the distance a vehicle must move to accelerate using a rail launcher, or the length of travel down a runway for a conventional takeoff.

 

L=|V|22a

 (m

 or ft

)

 

Where,

  • v
    – air vehicle velocity (m/s
     or ft/s
    )
  • a
     – acceleration of the vehicle (m/s2
     or ft/s2
    )

 

 

    1. Stroke Length Deceleration

 

Like the stroke length for acceleration discussed above, the equation to calculate the stroke length for deceleration is the negation of the previous equation. The negative ensures that we have a positive length despite the negative value for acceleration as the vehicle decelerates to a stop.

 

L=-|V|22a

  (m or ft)

 

Where,

    • v
      – air vehicle velocity (m/s
       or ft/s
      )
    • a
       –deceleration of the vehicle (m/s2
       or ft/s2
      )

 

    1. Energy Imparted from Launch

 

At launch, energy is imparted onto the vehicle for it to transition from an at rest state to a launched state. The following equation defines the relationship between energy imparted from common sources to the sum of kinetic energy, potential energy, and consumption of stored energy.

 

ΔELauncher+ΔEPropulsion+ΔELosses=WTOΔV22g+Δh+ΔEStored

 

 

Where,

  • ΔELauncher
     – energy imparted by the launcher (J or ft-lb)
  • ΔEPropulsion
     – energy imparted by the propulsion system (J or ft-lb)
  • ΔELosses
     – energy losses during launch event (J or ft-lb))
  • WTO
     – weight of vehicle at time of launch/takeoff (N
     or lb
    )
  • g
     – gravity (9.8 m/s2
     or 32.17 ft/s2
    )
  • ΔV
     – change in velocity (m/s
     or ft/s
    )
  • Δh
     - height difference from initiation of launch to height at completion of launch (m
     or ft
    )
  • ΔEStored
     – stored energy, e.g. fuel, consumed during launch event (J or ft-lb)

 

Dependent upon the launcher used, one or multiple of the ΔE

 components may not be applicable. For instance, a conventional takeoff does not utilize a launcher; therefore, ΔELauncher=0

. As another example, consider a rail launch system where the UA’s propulsion system is not active until some post-release velocity is attained, which would make ΔEPropulsion=0

.

 

 

Additionally, during preliminary design where energy losses are unknown, ΔLosses

 is often neglected.

 

 

Finally, if stored energy is not consumed during the launch event, ΔEStored=0

.

 

 

 

    1. Energy Absorbed for Recovery

 

ERecovery=WRecovery2g|ΔV|2+WRecovery⋅Δh

 

Where,

  • ERecovery
     – energy imparted from the recovery event (J
    )
  • WRecovery
     – weight of vehicle at time of recovery (N
     or lbf
    )
  • g
     – gravity (9.8 m/s2
     or 32.17 ft/s2
    )
  • ΔV
     – change in velocity (m/s
     or ft/s
    )
  • Δh
     - height difference from initiation of recovery to final resting height (m
     or ft
    )

 

    1. Simplified Wing Loading Estimate

 

When a vehicle is launched or recovered, the wing loading should be considered to ensure that the safety margin for peak loads is not exceeded.

 

The following equation estimates the wing loading in terms of the lift applied to the wing surface divided by the wing’s surface area. Manipulating the Lift equation presented in the aerodynamics and airframe design worksheet, the wing load can be solved as:

 

L/S=0.5⋅ρ⋅V2CL

  (N/m2

 or lb/ft2

)

 

 

Where,

  • ρ
     – air density (kg/m3
     or slug/ft3
    )
  • V
     – velocity at which the vehicle is traveling (m/s or ft/s)
  • CL
     – coefficient of lift given current air vehicle state (velocity and altitude)

 

 

Please note that this is not a precise measurement across various portions of the wing surface. Rather, it provides an estimate across the entirety of the wing.

 

 

Exercise:

 

  1. A hand launched UA weighing 10 lb attains a launch velocity of 15 kts. If the wing has an aspect ratio of 8 and a CL,Max=1.1:
    • What is the maximum wing loading?
    • What is the minimum wing area?
    • What is the wing span?

 

 

  1. Conventional

 

    1. Conventional Launch

 

The energy imparted during launch is provided by the propulsion system and not by a launch mechanism. Therefore, for the energy equation (see Section 1), ΔELauncher

 is set to zero (0).

 

 

 

Exercise:

 

  1. A conventional takeoff of an unmanned aircraft applies a 0.5 g acceleration to attain a 70 knot launch speed. At what distance must the air vehicle travel before it attains its launch velocity?

 

 

 

    1. Conventional Recovery

 

For recovery of a UAS using conventional landing gear, the energy equation can be rewritten as to account for energy expended for braking:

 

ΔEPropulsion+ΔELosses+ ΔEBraking=WTO-VApproach22g+Δh+ΔEStored

 

 

Where,

  • ΔEPropulsion
     – negative energy (i.e. imparted from reverse thrust) (ft-lb
     or J
    )
  • ΔELosses
     – energy losses during landing (ft-lb
     or J
    )
  • ΔEBraking
     – negative change in energy from nonthrust braking (brakes or spoilers)
  • WRecovery
     – weight of vehicle at time of recovery (N
     or lbf
    )
  • g
     – gravity (9.8 m/s2
     or 32.17 ft/s2
    )
  • VApproach
     – velocity at initiation of approach (m/s
     or ft/s
    )
  • Δh
     - height difference from initiation of landing to final resting height (m
     or ft
    )

 

  1. Rail

 

For rail launch systems, the UAS is pushed along a rail and released with a sufficient force that it can attain a desired launch velocity after release. Rocket, pneumatic, spring, and other mechanisms can be used to impart energy for the launch event.

 

    1. Release Velocity

 

Upon reaching the end of the rail, the system has attained is release velocity. This velocity is different from its launch velocity, which is the desired velocity for which the vehicle begins its climb-out. The following equation can be used to determine the release velocity

 

VRelease=|ΔV|2+2gΔh-ΔhRelease

  (m/s or ft/s)

 

 

Where,

  • VRelease
     – release velocity (m/s
     or ft/s
    )
  • ΔV
     – change in velocity from initial state to launch state
  • Δh
     - height difference from initiation of launch to height at completion of launch (m
     or ft
    )
  • ΔhRelease
     - height of initial release of the UA (m
     or ft
    )

 

    1. Rail – Acceleration and Deceleration Stroke Lengths

 

The rail is comprised of an acceleration section and a deceleration section (if a shuttle is used). The total length is the sum of the length of these two sections. Each length can be derived utilizing the stroke length equations for acceleration deceleration as specified above.

 

LRail=LAccel+LDecel

 (m or ft)

 

 

    1. Pneumatic Rail Launch

 

When a pneumatic launch mechanism is used, a pneumatic piston drives a shuttle along the rail.

 

The force delivered by the piston is equal to the:

FPiston=ΔPΠ4DPiston2

 (N

 or lb

)

 

              Where,

  • ΔP
     – pressure differential (N/m2
     or psi)
  • DPiston
    - diameter of the piston (cm2
     or in2
    )

 

The force delivered by the shuttle to launch the vehicle can be calculated as follows:

FShuttle=WTO+WShuttleag+sinθlauncher

              (N

 or lb

)

 

Where,

  • WTO
     – aircraft takeoff weight (N
     or lb
    )
  • WShuttle
     – weight of shuttle (N
     or lb
    )
  • a
     – acceleration of the aircraft along rail (m/s2
     or ft/s2
    )
  • θlauncher
     – elevation angle of the launcher (degrees
    )

 

The piston stroke length corresponds to the mathematical relationship for acceleration stroke length. That equation can be used to derive unknowns if the stroke length is already known. Given the ratio of piston length to acceleration length, the following relationship between the Force delivered by the shuttle given the force delivered by the piston exists.

 

FShuttle=FPistonLPistonLAccel

 

 

Where,

  • FPiston
     – force delivered by the piston (N
     or lb
    )
  • LPiston
     – piston stroke length (m
     or ft
    )
  • LAccel
     – rail acceleration length (m
     or ft
    )

 

 

 

Exercise:

 

  1. A pneumatic launch of an unmanned aircraft applies a 10 g acceleration to attain a 70 knot launch speed. At what is the acceleration stroke length of the air vehicle to it attains its launch velocity?

 

  1. A rail-based pneumatic launch system has an elevation angle of 30 degrees, a total rail length of 20 ft, and an acceleration stroke length of 15 ft. The UA is installed at 5 ft above ground level and must clear a 50 ft obstacle and attain a launch velocity of 60 kts to begin climb out once it has cleared the obstacle.

 

    • What is the release velocity?
    • How much energy is imparted to the UA through the launch event?
    • What is the average acceleration of the vehicle?
    • What is the deceleration of the dolly?

 

  1. For the system described in Exercise 5, if a shuttle weight 50 lb is used with a piston stroke length of 5 ft and a piston diameter of 6 in.
    • What is the force applied to the shuttle during acceleration?
    • What is the piston force?
    • What is the pressure differential of the piston?

 

 

  1. Rocket Launch

 

For a rocket launch, a variable rate of Thrust (N or lb) propels the rocket toward a desired launch velocity. The total impulse is described in the textbook in equation 11.41 as the integral of thrust over time through the duration of the launch event.

 

The total impulse of the rocket launch relates to the change in momentum as described in the following equation.

 

ITot=WTOg⋅ΔV

 (Ns

 or lb s

)

 

Where,

  • WTO
     – takeoff weight (N
     or lb
    )
  • g
     – gravity (9.8 m/s2
     or 32.17 ft/s2
    )
  • ΔV
     – change from at rest velocity to launch velocity (m/s
     or ft/s
    )

 

From the above equation, you can solve for the change in velocity resulting from the rocket launch if the total impulse is known.

 

 

One advantage of a rocket-based launch mechanism is it can achieve a zero-length takeoff distance. To achieve this goal, the vertical component of the thrust vector must exceed the takeoff weight of the vehicle and the takeoff weight of the rocket. This relationship is expressed as follows:

 

TRocketsinθRocketWTO+WRocket

 

 

Where

  • TRocket
     – thrust delivered by the rocket (N
     or lb
    )
  • θRocket
     – angle of the rocket launch vector (degrees)
  • WTO
    – takeoff weight (N
     or lb
    )
  • WRocket
     – weight of rocket (N
     or lb
    )

 

 

Exercise:

 

  1. If a rocket launcher produces a total impulse of 2,000 N.s to attain a velocity of 100 m/s from its initial at rest state, what is the maximum launch weight?

 

 

 

  1. Tension Line Launch

 

The tension-line launch technique supports the launch of small unmanned aircraft systems.

 

The textbook presents the general equation for the transfer of energy from the tensioned line to the vehicle in equation 11.48 as the integration of the tension of the cable as a function of its stretch distance from its starting point.

 

This relationship simplifies for spring and elastic chord tension lines utilizing the spring constant, k

. This equation is given as:

 

 

ΔELine=12k?⋅ΔX2

 

 

Where,

  • k
     – spring constant for the tensioned line (N/m
     or lb/ft
    )
  • ΔX
     – change in tension lined link from its rest state to its fully stretched state (m
     or ft
    )

 

Using the above equation for ΔELine

, additional unknowns can be derived using the impulse-momentum equation for launch events.

 

 

  1. Horizontal Arresting Cable Recovery

 

Within the textbook, a horizontal arresting cable is described as a type of arresting cable used for recovery. In the ideal case, the aircraft snags the arresting cable such that the tension applied symmetrically between the two sides of the arresting cable as the UA continues to move forward in the x-axis.

 

The forces and distance traveled to arrest the vehicle can be calculated as a step-wise analysis updating the distance traveled and the force of the cable in the x-axis utilizing the following equations.

 

  1. The resulting force in the x-axis as a function of the current angle of the cable relative, θCable
    , to its resting state is opposite the forward force of the vehicle and can be calculated as:

 

Fx,Cable=-2TCablesinθCable

   (N

 or lb

)

 

 

Where

  • TCable
     – tension of cable (N
     or lb
    )
  • θCable
     – angle of cable relative its initial rest condition (degrees)

 

 

  1. The cable angle can be calculated as the arctan of the distance traveled in the x-axis past the cable’s resting point.

 

θCable=tan-1ΔXAV1/2 ⋅ LCable

                 (degrees)

 

Where

  • ΔXAV
     – distance of air vehicle relative to the arresting cable’s at rest position in the x-axis (m
     or ft
    )
  • LCable
     – distance between the arresting cable pullies (m
     or ft
    )

 

  1. The vehicle’s acceleration (negative because of deceleration) for a particular time step can be calculated using the following equation:

 

a=Fx,Cable  ⋅gWRecovery

  (m/s2

 or ft/s2

)

 

Where,

  • Fx,Cable
     – arresting force in the x direction applied to the vehicle (N
     or lbf
    )
  • WRecovery
     – recovery weight (N
     or lbf
    )
  • g
     – gravity constant (9.8 m/s2
     or 32.17 ft/s2
    )

 

  1. The distance traveled by the air vehicle between time steps.

 

ΔXAV,i=Vi-1⋅ΔT+12ai⋅ΔT2

 

 

  • Vi-1
     – velocity at time step  i-1
     (m/s
     or ft/s
    )
  • ΔT
     – duration of time step
  • ai
     – acceleration of vehicle at time step i
    . (m/s2
     or ft/s2
    )

 

Utilizing the above equations, the stop distance and stop time can be estimated. Additionally, the loads on the vehicle can be estimated by calculating Fx,Cable.

 

 

Exercise:

 

  1. Using a step-wise analysis of a horizontal arresting cable system with tensioners located 50 ft apart using a time step of 0.2 s, determine the stopping distance and time to stop for a 750 lb UA landing at a velocity of 100 kts when it snags the arresting cable.
    • For each time step, determine the vehicle’s velocity, acceleration, and distance traveled from initial snag of the cable until the vehicle is at rest.
    • Note the final stopping distance and time.

 

  1. If the system integrators determine that a maximum stopping distance of 50 feet is required for the UA for the system described in exercise 8, what is the cable tension required to meet this requirement?

 

 

  1. Parachute Recovery

 

Parachute recovery permits both gliding and non-gliding descent profiles. This section briefly highlights the calculations required for a non-gliding UAS descending at the chute’s terminal velocity, VT

.

 

 

The terminal velocity is attained when the drag produced by the chute equals the recovery weight of the aircraft.

 

WRecovery=DChute=12⋅ρVT2CD,ChuteSChute

            (N

 or lb

)

 

Where,

  • D
     – drag produced by the parachute (N
     or lb
    )
  • ρ
     – air density (kg/m3
     or slug/ft3
    )
  • CD,Chute
     – coefficient of drag for the parachute (unitless)
  • SChute
     -surface area of the parachute (m2
     or ft2
    )

 

 

The surface area of a round chute relative to its diameter is equal to:

SChute=π4DChute2

 

 

Where,

  • DChute
     – diameter of the chute (m
     or ft
    )

 

Utilizing the above equations, the diameter of the parachute can be calculated as:

DChute=8WRecoveryπ⋅ρ⋅VT2CD,Chute

 

 

Where,

  • WRecovery
     – air vehicle’s recovery weight
  • ρ
     – air density (kg/m3
     or slug/ft3
    )
  • CD,Chute
     – coefficient of drag for the parachute (unitless)
  • VT
     – terminal velocity of air vehicle under chute (m/s
     or ft/s
    )

 

 

Exercises:

 

  1. An unmanned aircraft weighing 1,000 lb at the initiation of recovery is recovered utilizing a round parachute with a drag coefficient of 1.1. Given a descent rate of 20 ft/s, what is the diameter of the parachute?

 

 

 

  1. Impact Attenuation

 

Section 11.17 describes impact attenuation if the vehicle were permitted to impact the ground as part of its arresting technique. If the vehicle is designed with crumple zones, the impulse-momentum equation can be described as:

 

FZLStroke=WRecovery2gVz,Impact2

 

 

Where,

  • FZ
     – force of impact (N
     or lb
    )
  • LStroke
    - stroke length of vehicle coming to rest upon impact (m
     or ft
    )
  • WRecovery
     – weight of vehicle at recovery (N
     or lb
    )
  • Vz,impact
     – velocity of vehicle at impact (m/s
     or ft/s
    )

 

From this equation, the level of force of the impact, FZ

, along the z-axis can be derived based upon the length of the vehicle’s crumple zone, LStroke

 along the z-axis and the vehicle’s velocity along the z-axis, Vz ,   Impact

.

 

 

 

Exercise:

 

  1. A 10 lb UA has a nose crumple zone of 12 in in length. If the vehicle is oriented for a vertical surface impact at 150 mph, what is the average force of the impact event?
  2. For the above UA, if the vehicle’s speed is reduced to 30 mph using a drogue chute, what is the average force of the impact event?     

Option 1

Low Cost Option
Download this past answer in few clicks

26.99 USD

PURCHASE SOLUTION

Already member?


Option 2

Custom new solution created by our subject matter experts

GET A QUOTE

Related Questions