Reference / Paper · 2017
Peak of Flight #456: A Guide to Optimal Altitude: Part 2
Apogee Components' Peak of Flight newsletter, issue 456 (2017-11-14).
- peak-of-flight
- apogee-components
- model-rocketry
Reference / Paper · 2017
Apogee Components' Peak of Flight newsletter, issue 456 (2017-11-14).
Apogee Components' Peak of Flight newsletter, issue 456 (2017-11-14).
ISSUE 456 November 14th, 2017 IN THIS ISSUE A Guide to Optimal Altitude: Part 2 www.ApogeeRockets.com/Rocket-Kits/Skill-Level-3-Model-Rocket-Kits/Sandia-Sandhawk-13mm A Guide to Optimal Altitude: Part 2 This material copyrighted 2017, ARA Press. Reprinted by permission. Where m is the mass of the propellant exhausted (one brick is ten pounds) 4.54 kg, and V is the velocity of the throw (Clarke assumed 20 mph) 8.94 m/s. By Steve Ainsworth Chapter 3: Rocket Fundimentals This chapter provides some equations that we will Lets assume the mass of the trolley is 200lbs (90.7 kg). need later and is a good review of the basics. I recently What mass of bricks must be carried for the final velocity of read a book titled The Exploration of Space by Arthur C. the trolley to be equal to the throwing velocity (8.9 m/s)? If Clarke (published in 1951) which provided a determination your first guess is 200 lbs, consider that the first brick thrown of the maximum velocity possible utilizing a rocket motor must accelerate not just the trolley but also the other bricks. when starting from earth. Since Clarke is one of my favorWe need a total brick mass equal to 1.7 times the mass of ite authors, I bought a used hardbound copy from Amazon the trolley or 344 lbs (156 kg) of bricks (Figure 3-1, data books for just over $5! When the book arrived, I remempoint 1). bered reading it as a teen from the school library, Table 3-1 and relearned some rocket fundamentals. 1 Throwing Bricks From a Trolley Car 2 3 4 --- Basic Units --Time Length Mass Metric mks 1.0 0.305 4.536 Symbol s m kg Units second meter kilogram Basic Units To English * 1.0 * 3.281 * 2.205 5 6 7 gravity velocity force 8 Final Velocity The analysis presented by Clarke is very 9.807 8.940 1,431.10 32.19 g v nt (or N) v straightforward and lends itself to an intuitive m/s/s m/s newton km/hr understanding of rocket fundamentals without m/s^2 m/s kg-m/s^2 having to gnaw through difficult math concepts. * 3.281 * 3.281 * 0.225 * 0.621 As a matter of fact, Clarke’s analysis helped me English fp 1.0 1.0 10.0 32.17 29.3 321.7 20.0 understand several non-intuitive (for meFigure at 1least) Symbol s f lbm (or lb) g v lbf (or lb) v concepts of rocketry like Specific Impulse (Isp) Units second feet pounds f/s/s f/s pounds miles/hr which has units of just seconds, and why a high Basic Units f/s^2 f/s lbm-f/s^2 To Metric * 1.0 * 0.305 * 0.454 * 0.305 * 0.305 * 4.448 * 1.61 exhaust velocity for a rocket motor is a very good thing. The mathematical analysis itself Figure 3-1 originated with Konstantin Tsiolkovsky. constant exhaust velocity (8.9 m/s) and final mass (90.7 kg) Pavel Checkov would be proud. 24 The thrust of a rocket is a result of 22 the conservation of momentum and a momentum exchange between the 20 rocket and the expelled propellant. | 18 17.8 ‐‐‐‐‐‐ Consider a man standing on a fric| tionless (to keep it simple) railway trol16 ley car that has a stack of ten pound 14 bricks on it. In the following discus12 sion, “trolley” will mean the trolley car plus the man. When the man throws 10 | a brick off the back of the trolley, the ‐‐‐‐‐‐ | 8 trolley moves in the opposite direction 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 in accordance with the conservation Propellant Mass of momentum. The momentum added to the brick by the man throwing it rearward, is added to a If 344 lbs of bricks are thrown off at 20 mph (8.94 m/s), forward motion of the trolley. The math (see Table 3-1 for then the velocity of the trolley after the last brick is thrown off unit conversions) goes like this: will be 20 mph. What throwing velocity is needed to double the final velocity of the trolley? Intuition says the throwing Propellant Momentum P (kg-m/s) = m (kg) *V (m/s) velocity must double. That is correct. If 344 lbs of bricks are thrown off at 40 mph (17.9 m/s) then the velocity of the trolley after the final brick is thrown off will be 40mph (Figure 3-2, Page 3). About this Newsletter You can subscribe to receive this e-zine FREE at the Apogee Components website www.ApogeeComponents.com, or by clicking the link here Newsletter Sign-Up Page 2 Newsletter Staff Continued on page 3 Writer: Steve Ainsworth Layout/Cover Artist: Chris Duran Proofreader: Will Franks Issue 456 | November 14th, 2017 A Guide to Optimal Altitude: Part 2 Continued from page 2 Can the man get the trolley going faster by adding more bricks instead of throwing them faster? Yes, but to double the final velocity we must have a mass of bricks equal to 6.4 times the mass of the trolley (Figure 3-1, Page 2). That is, for the final velocity of the trolley to double with the same throw (exhaust) velocity, we need to have 1,280 lbs (580.6 kg) of bricks. In order to double the final velocity, we can either increase the propellant exhaust velocity by a factor of 2.0 or increase the propellant mass by a factor of 6.4. The exhaust velocity for a typical high powered rocketry motor is on the order of 6,000 f/s (1,829 m/s or 4,091 mph). The Figure 2 energetic liquid-fueled motors can provide an exmore haust velocity of 7,000 f/s (2,134 m/s or 4,773 mph) (Ref 9). The velocity required to achieve orbit is on the order of 26,400 f/s (8,047 m/s or 18,000 mph). In the following equation: g = gravity acceleration = 32.2 f/ s2 = 9.807 m/s2 F = Thrust magnitude nt (kg-m/s2) Isp = F/(m * g) = Specific Impulse [kg-m/s2]/ [(kg/s)*(m/s2)] = seconds Isp is defined with units of seconds so that is has the same value regardless of the units system used for its calculation. So Isp in Metric or English units will have the same value. The thrust over mass flow rate is divided by g to eliminate the mass and length units so that Isp will be the same in both English and Metric units. The result is that the units for Isp is seconds. For rocket motors Equation 3.1:, F/m = Ve (1). Where Ve is the exhaust velocity, or for our trolley, the velocity of the thrown bricks. So we get: Equation 3.2 : Isp = Ve/g. Isp is a function of exhaust velocity. The higher the exhaust velocconstant propellant mass (156 kg) and final mass (90.7 kg) ity, the higher (and better) the Isp. Clarke’s discussion above shows why a high exhaust velocity is VERY desirable for rocket motors. Figure 3-2 65 60 55 50 Final Velocity m/s This material copyrighted 2017, ARA Press. Reprinted by permission. 45 40 Ten Pound Brick Propellant 35 30 25 17.8 | | 20 15 10 | ‐‐‐‐‐‐‐‐‐ | 5 0 0 5 10 Specific Impulse 15 20 25 30 35 40 Exhaust Velocity m/s People that design rocket motors use what is termed Specific Impulse (Isp) as a performance parameter to compare one motor system with another. Isp is the thrust force as a function of the propellant mass flow rate (m) kg/s. 45 Lets take a look at Clarke’s brick propelled trolley again. For the 20 mph (= 29.3 f/s) exhaust velocity, the Isp = [29.3 f/s]/[32.2 f/s2] = 0.91 seconds. If we could get the guy to throw 100 mph bricks, the Isp would be 4.6. Table 3-2 shows how 50 55 60 65 it compares to actual rockets. Looks like the brick propellant system is a real brick. To do or not to do the math Clarke did not do the math. The math requires just a bit of calculus. For the record, here’s the math. You can skip down to Equation 3.3 (Page 4), the Ideal Rocket Equation, if you would like. Continued on page 4 Gyro Chaser Helicopter Rocket • Unique ‘transforming’ rocket - looks like a normal rocket, but then rotor blades pop out at ejection • Competition efficiency: high flights and long descent time • Features curved rotor blades and free-spinning hub, just like those used in international competitions • Versatile: can use any 18mm diameter motor • Comes with video instructions for error-free assembly www.apogeerockets.com/Rocket_Kits/Skill_Level_4_Kits/Gyro_Chaser Page 3 Issue 456 | November 14th, 2017 A Guide to Optimal Altitude: Part 2 This material copyrighted 2017, ARA Press. Reprinted by permission. Continued from page 3 Momentum = mv; mass times velocity (definition) dm = infinitesimal (almost zero) propellant mass to be exhausted m = mass of the vehicle including propellant mass dm (kg) Mi = mass initial = m (kg) Mf = mass final = (m-dm) m = mass flow rate (kg/s) dv = infinitesimal change in velocity of the vehicle Ve = exhaust velocity of the propellant mass (how fast a brick is thrown) (m/s) Delta v = change in velocity of the vehicle (m/s) ln = natural log Table 3-2 ISP Values Type Brick Cold Gas Liquid Solid Hybrid Nuclear Electric low Isp 0.9 60 140 160 160 800 500 high 4.6 250 460 300 350 6,000 10,000 (m-dm)dv = -dm*Ve (momentum of the vehicle equals the momentum of the propellant; conservation of momentum) = mdv + dm*dv = -dm*Ve Given; dm* dv = 0 Then: mdv = -dm*Ve, solving for dv; dv = -Ve * dm/m and after integrating Delta v = -Ve*(ln(Mf) – ln(Mi)) = -Ve*ln(Mf/ Mi) Equation 3.3 : Delta v = -Ve*ln(Mf/Mi); This is the formula Clarke used to get the mass of bricks required for a given final velocity and a given exhaust velocity. It is known as the Ideal Rocket Equation. Equation 3.4: F = m * Ve (kg-m/ s2) Actually F/m = effective exhaust velocity which differs from our use of Ve for the brick example slightly. For an actual rocket motor the effective exhaust velocity is equal to the exhaust velocity plus the pressure thrust and reduced by the nozzle efficiency. Continued on page 5 Join Tripoli.org Mention Apogee Components https://www.apogeerockets.com/TARC_Supplies Page 4 Issue 456 | November 14th, 2017 A Guide to Optimal Altitude: Part 2 This material copyrighted 2017, ARA Press. Reprinted by permission. Continued from page 4 Chapter 4: Optimal Velocity Most hobbyists are aware that there is an optimum weight for a rocket with a given motor that will cause the rocket to achieve the greatest altitude. Some flight simulation software will even calculate that optimum weight for you. Is there also an “optimum velocity” that will provide the greatest altitude for a given motor impulse and airframe? Put simply, yes. But it is not that simple. The “optimum velocity” varies with altitude. It is time to go metric. English units have a problem for rocketry use. In the everyday world we refer to mass by the term lbs. In English units, the term for mass is actually Slugs (or even lbs. mass vs. lbs. force), which is lbs./g (g = force of gravity). This gets very confusing. In the mks (meters kilograms seconds) metric system, force is newtons (n) and mass is kilograms (kg). I have included a conversion table (Table 3-1, Page 2) for those that still want the final numbers in English units. Lets Review a Bit If a rocket has an empty mass of 20kg, what mass of propellant is required for it to achieve a final velocity of 1000m/s (2237mph) if the exhaust velocity for the motor is 2000m/s? From the Ideal Rocket Equation (Equation 3.3, Page 4 and Equation 4.1 below), we find that the rocket would need 0.65 * 20kg or 13kg of propellant with an exhaust velocity of 1000 m/s to achieve that final velocity. Recall that for an exhaust velocity of 2000m/s, the propellant would have an Isp of (2000 m/s) / (9.8 m/s2) = 203.9 seconds. This would be a motor with a total impulse of 26,000 n-s (13kg * 2000 m/s). Equation 4.1: Ideal Rocket Equation (solved for initial mass): mi = mf * e(Vf / Ve) mi = mass initial (rocket + propellant) mf = mass final (20kg) Chapter 4 Optimal Velocity 4 Vf = Final Velocity (1000 m/s) assuming a zero initial velocity. Ve = Exhaust Velocity (2000m/s) This calculation assumes that the rocket is not acting against gravity or drag (it is in space). In the real world of High Powered Rocketry, neither of these assumptions is accurate. When a rocket is rising from the earth, it must overcome gravity and the drag forces of the air. The force of gravity subtracts 9.807 m/s from the rockets vertical velocity for each second of flight! If the motor has a burntime of 20 seconds, gravity reduces the final burnout velocity by 196 m/s. If it had a burnout velocity of 1000 m/s without gravity, you would see a burnout velocity of 803 m/s (= 1000 – 20 *9.807). If the rocket had a 2 second burntime, the burnout velocity would be 980 m/s (= 1000 – 2*9.807). Therefor, the sooner you get the rocket to its maximum velocity (the shorter the motor burntime), the less time gravity has to reduce the total force of the motor slow it down. However, shortening the burntime causes another problem to crop up. As shown above, the shorter the burntime, the higher the maximum velocity of the rocket. Unfortunately, as the rocket travels faster in the air, the aerodynamic drag force increases exponentially. Equation 4.2: D = 0.5 * Cd * Rh *v2 *A D= Drag Force (Newtons) Cd = Drag Coefficient (Usually between 0.01and 0.5) (Unitless) Rh = Air Density (kg/m2) v = Rocket Velocity (m/s) A = Reference area of the airframe (m2) Continued on page 6 https://www.apogeerockets.com/Rocket-Kits/Skill-Level-3-Model-Rocket-Kits/Slo-Mo Page 5 Issue 456 | November 14th, 2017 A Guide to Optimal Altitude: Part 2 This material copyrighted 2017, ARA Press. Reprinted by permission. the air density (Rh), and Rh varies with altitude. We have too many things varying to make it nice and neat. My approach is to run the calculations at specific increments of velocity and altitude and add gravity losses to drag losses to find the velocity that provides the minimum total losses. Continued from page 5 Apogee Feet Roughly speaking, within the atmosphere, the drag force increases with the square of the velocity. If you double the velocity, the drag force Figure 4-1 increases by a factor of 4! The longer the bur time, the great105,000 er the impact of gravity (and gravity 40,000 n-s Motor turning). 100,000 The shorter the burn time, the greater the impact of drag. 95,000 The logical question is for a given rocket, what motor burntime (average thrust) will yield the highest altitude? 90,000 I ran about 20 RockSim simulations using the same airframe and total 85,000 impulse motor with a range of burntimes, to see if there would be an optimum burntime. Optimum being the 80,000 burntime that provides the highest altitude for a given total impulse. The 75,000 results of those runs are provided in Figure 4-1. 70,000 Typically with rocketry, we are try0 5 10 15 20 25 ing to get the highest altitude possible Burntime Seconds from a particular total impulse motor and airframe design. So, if you could Figure 4-2 have any thrust curve that you wanted, what thrust curve would provide the highest altitude for 1.0 a given airframe? To find this out, we need to determine 0.9 the optimum velocity for that airframe as it ascends. We need to calculate the gravity losses and add those 0.8 to the drag losses for a range of velocities at each altitude 0.7 increment. Then by inspection, we can discover the velocity that will yield the smallest losses, and thus the highest 0.6 altitude. This information together with the airframe design 0.5 data will allow us to calculate the thrust required at each interval, giving us the optimum thrust curve. 0.4 Equation 4.3: Lgv = 9.807 m/s2 * Time = Lgv m/s = Veloci0.3 ty loss due to gravity 0.2 The gravity loss is just 9.807 m/s/s times time, which provides the velocity loss m/s. For this analysis, we are 0.1 using an altitude increment of 1,000 meters. The time 0.0 required for the rocket to travel that increment is = 1000m / 0.0 0.5 1.0 1.5 2.0 2.5 Vm/s = 1,000/V seconds. Mach (Mach 1.0 = 340 m/s) The drag loss is provided by Equation 4.2 (Page 5). The drag calculation requires that we know the drag coefficient Cd. The problem here is that Cd is not a constant (Figure 4-2). It varies with the rocket airspeed as a function of the Mach number (Mach 1 = the speed of sound = 340m/s at ground level with a standard atmosphere). The actual airspeed of Mach 1 varies with altitude (Figure 4-3, Page 7), as its a function of air density and temperature. The drag loss calculation also requires that we know 35 40 Cd 30 Page 6 3.0 3.5 4.0 Continued on page 7 Issue 456 | November 14th, 2017 A Guide to Optimal Altitude: Part 2 This material copyrighted 2017, ARA Press. Reprinted by permission. Continued from page 6 1.3 1.2 Speed of Sound 1.1 100k 1.0 Air Density 0.7 0.6 0.5 0.4 0.3 0.2 0.1 15,000 20,000 Altitude Meters 25,000 30,000 0.0 40,000 35,000 Grav Booster at 5,000 meters Drag 450 400 350 Velocity m/s 300 250 Sum 200 150 100 50 n - s loss per 100 meters 0.8 kg per cubic meter 0.9 meters per second Another approach is to simply use Figure 4-3 RockSim and run a range of burn355 times. 350 Figure 4-4 shows the gravity 345 loss, drag loss and summed loss340 es for an airframe that consists of 335 330 a booster connected to a second 325 stage (sustainer) for a range of ve320 locities at a specific altitude (5,000 315 meters). Figure 4-5 provides the 310 data for the Sustainer at 20,000 me305 ters. After this data is plotted for alti300 tude increments up through 40,000 295 meters, and the optimum velocity is 290 0 5,000 10,000 picked out for each altitude increment we get Figure 4-6, which now includes the optimum velocities for Figure 4-4 both the booster with the sustainer, 400 and the sustainer by itself. 350 This graph confirms the ear300 lier finding that long burn motors 250 providing slower rockets go higher (as long as they go straight up). But 200 it also tells us just what velocity is 150 best. Now, if you run these calcula100 tions for your very sleek minimum 50 diameter rocket, find a motor that will provide velocities near the 0 optimum, launch on a calm day and are lucky enough to achieve a near vertical flight, then you can achieve the maximum possible altitude using that airframe and total impulse. These graphs are based on the to 100k rocket project described in Chapter 5. We will discuss more on this topic in the upcoming Peak of Flight Newsletter “A Guide to Optimal Altitude: Part 3” Continued on page 8 Need Rail Buttons And Stand-Offs? www.apogeerockets.com/Building_Supplies/Launch_Lugs_Rail_Buttons/Rail_Buttons www.apogeerockets.com/Rocket-Kits/Skill-Level-2-Model-Rocket-Kits/SkyMetra Page 7 Issue 456 | November 14th, 2017 This material copyrighted 2017, ARA Press. Reprinted by permission. A Guide to Optimal Altitude: Part 2 Continued from page 7 About The Author: Steve Ainsworth is a Civil Engineer with a Master’s degree in Physics from UNLV. He has published articles in both HPR and ER since he became a “Born Again Rocketeer” in 1994. Steve joined Tripoli Vegas in 1995 and began flying his “gizmo” rockets. In 1995, Las Vegas was the perfect setting for re-entering rocketry with Gary Rosenfield of Aerotech providing motor tutoring and Tom Blazinin (TRA # 003) as the local Prefect, coaching and signing for Steve’s Level 1 and Level 2 flights. While on an extended engineering assignment in northern California, Steve had evening time to think about rocketry. He used that time to learn enough aerodynamics to develop a computational method for flight predictions based on Gordon Mandell’s MIT Thesis articles. Join The NAR.org Mention Apogee Components 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 S t ustainer at 20,000 meters t Grav Drag 1,800 1,700 1,600 1,500 1,400 1,300 Velocity m/s 1,200 1,100 1,000 900 800 700 600 500 400 300 Sum 200 n - s loss per 100 meters Figure 4-5 1,800 1,700 1,600 1,500 1,400 1,300 1,200 1,100 1,000 900 800 700 600 500 400 300 200 100 0 Optimum Velocity Mach 5 Mach 4 Mach 3 Mach 2 Booster Sustainer Mach 1 35,000 32,500 30,000 27,500 25,000 22,500 20,000 17,500 15,000 12,500 10,000 7,500 5,000 2,500 100k | Feet 0 Meters per Second Figure 4-6 Altitude Meters Page 8 Issue 456 | November 14th, 2017