Reference / Paper · 2012
Peak of Flight #323: How To Compute Delay Time And Altitude Using Quadratics and Euler's Number
Apogee Components' Peak of Flight newsletter, issue 323 (2012-10-09).
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- apogee-components
- model-rocketry
Reference / Paper · 2012
Apogee Components' Peak of Flight newsletter, issue 323 (2012-10-09).
Apogee Components' Peak of Flight newsletter, issue 323 (2012-10-09).
In This Issue How To Calculate a Rocket’s Delay And Altitude Using Quadratics and Euler’s Number Cover Photo: New Way Rocketry’s Skyrise Kit. Get one at: www.ApogeeRockets.com/Rocket_Kits/Skill_Level_3_Kits/Skyrise Apogee Components, Inc. — Your Source For Rocket Supplies That Will Take You To The “Peak-of-Flight” 3355 Fillmore Ridge Heights Colorado Springs, Colorado 80907-9024 USA www.ApogeeRockets.com e-mail: [email protected] Phone: 719-535-9335 Fax: 719-534-9050 ISSUE 323 OCTOBER 9, 2012 Calculating the Delay Period Using Quadratics and Euler’s Number By Ian Voss {Editor’s Note: Before there was computer software that made our lives so easy, modelers had to calculate by hand the flight parameters of the rocket. This article shows you how to do just that. It is not likely that you’ll do these formulas by hand, but you’ll find this to be a good leap into rocketry eduction. I’m sure a few teachers will like their students to walk through this mathematical derivation. is only approximate so it is more practical to just base our calculations on the airframe diameter. Also, our author is from Australia, so you may notice some words spelled a little differently than we spell them here in the United States of America.} 4. You will need to know the weight of the rocket, i.e. the weight of the rocket when it is fully prepped and ready to launch. You could allow for propellant mass and use the average weight of the rocket but I won’t be doing this in my example. It would unnecessarily complicate what is only an approximation anyway. Simulation software is great for designing model rockets and determining how they will perform. I use and recommend them, especially if you are scratch-building. It takes the guesswork out of selecting a delay period (and a great many among other things) so that you can avoid high speed deployment events and potential damage to the rocket. On occasion though, you may find yourself without access to a computer (or the required software) wanting to test a previously untried combination of rocket and motor. For those in this situation, it’s possible to calculate the delay period with reasonable accuracy using only a pen and paper (although, having a calculator will definitely make the process easier). If you don’t use simulation software (I can’t imagine why you wouldn’t want to), these calculations can be set up on a spreadsheet. Before you get too excited though, there are some provisos: 1. This method does not account for dynamic stability and assumes that you are launching vertically in dead calm conditions. It does however, allow for aerodynamic drag. 2. Because we are allowing for drag, you will need to either know or take a guess at the CD (Drag Co-efficient). You will also need to know the maximum airframe diameter. You could calculate the projected frontal cross-sectional area including the fins and launch lug but it’s really not necessary. The fins and launch lug have a very small frontal area compared to the airframe and our ‘averaged’ CD About this Newsletter 5. This method is fine for low powered rockets. I’ve compared the values obtained from this method with what comes out of a simulator for several small rockets and they’re well within 10%. I have not tested for and would not expect accuracy for higher speeds and altitudes. Perhaps the reader could test the accuracy with their own measurements? 6. If your rocket has apogee detection, the delay period is irrelevant. You might however, want to use this method to determine the altitude at apogee. 7. Metric units will be used (metres, kilograms, Newtons) so if you are not familiar with these, now would be a good time to get familiar with them. With metric units, conversion factors can be avoided which simplifies the process. If you prefer to measure in feet, it’s easy enough to convert the results. 8. I would expect that finding the way through this process would be daunting for some. I would recommend taking a notepad with a template already in place if you intend to use this method at the range. An Overview Of The Process Here is an outline of the process. Armed with the above information, we’ll form a quadratic equation which will give Continued on page 3 Newsletter Staff You can subscribe to receive this e-zine FREE at the Apogee Components web site (www.ApogeeRockets.com), or by sending an e-mail to: [email protected] with “SUBSCRIBE” as the subject line of the message. Page 2 3. You will need to know the Total Impulse and the Thrust Duration of the motor you intend to use. This information should be supplied with the documentation that comes with the motor. Or, you could take values from the motor manufacturer’s catalogue. ISSUE 323 Writer: Tim Van Milligan Layout / Cover Artist: Tim Van Milligan Proofreader: Michelle Mason OCTOBER 9, 2012 Continued from page 2 Calculating the Coast Delay Period us the maximum velocity (Vmax) of the rocket (or a close approximation at least). Once we know Vmax , we can form another quadratic (or something that closely resembles one) and use that to determine the delay period. There’s some calculus along the way (integration) but I’ll avoid the usual notation and instead try to make it look like a template where you can just plug in the values. As a bonus, once the delay period is known, a few more simple calculations will give us the altitude at apogee. The critical value is CD and one would do well to keep in mind that the value we choose may not be a true value as might be obtained from wind tunnel testing. Rather, for the purpose of these calculations, it is more like an average value. Remember that we are not allowing for dynamic stability or weather conditions so what may be correct for one flight may give erratic values with different motors and/ or different conditions. When using different motors, the static margin will change and this will have an effect on the dynamic stability. As the dynamic stability changes, the overall effect of aerodynamic drag is likely to change. To allow for dynamic stability and weather conditions, the only practical solution would be to use simulation software which brings us back to the reason for this article. In short, this method is aimed at providing an approximation of how a model rocket will perform with a given motor when you don’t have access to a computer. The best we can hope for are ‘ballpark’ figures – not as good as a simulation but still better than a complete guess. Calculating the Maximum Velocity Rocket motors are designed to produce their maximum thrust as quickly as possible so that the rocket reaches a stable speed before it leaves the launch rod. While this is good in terms of providing a stable flight, it makes it rather complicated to analyse what is happening to the rocket over a series of short time increments. As the thrust changes, so does the acceleration. The mass of the rocket is also changing as the propellant is burnt. The only constant is gravity! Luckily for us, initially we won’t be looking for any value other than Vmax and we can find this by making some assumptions. Step 1 is to ignore the motor’s thrust curve entirely and simply assume that the thrust is constant throughout the thrust phase. This simplifies the calculations considerably which is what we want for a pen and paper (and calculator) exercise. In my example, I’m going to use an Estes ‘Alpha’ fitted with a A8-3 motor and a lift-off weight of 50 grams. For the A8-3 motor, the Estes catalogue specifies a total impulse of 2.50 Newton.seconds (Ns) and a thrust duration of 0.5 seconds. Total Impulse: ITotal = 2.50Ns Thrust Duration: t1 = 0.5s The average thrust for the ½ second interval is: FT = ITotal / t1 = 2.50 / 0.5 = 5N We’ll need this value soon. Step 2 is to find the weight of the rocket. This is easily determined using Newton’s 2nd Law (F = ma). A common misconception is that the unit for weight is kilograms. From a strictly scientific perspective, weight is the result of the effect of gravity and is a force and force is measured in Newtons. Note that we will define everything in the upward direction as having a positive value and everything in the downward direction as having a negative value. Continued on page 4 Rocket Jewelry • The Perfect Launch-Range Accessory • Subtle, Tasteful, Fashionable & Distinctive • Makes A Great Gift for Family and Friends • Display Your Passion for Rocketry www.ApogeeRockets.com ISSUE 323 OCTOBER 9, 2012 Page 3 Continued from page 3 Calculating the Coast Delay Period ‘Alpha’ uses BT50 body tube which has a diameter of 24.8mm. The mass of our rocket is: m = 0.050kg The acceleration due to gravity is: ag = -9.8 m/s2 This is an estimate of the rocket’s drag Cd = 0.550 co-efficient. Remember that this is more like an average value for our calculations. Therefore, the weight of the rocket is: FW = (0.050kg)(-9.8m/s2) v1 = ? This is the velocity of the rocket which is as yet undetermined. = -0.5N We’ll also need this value. It’s interesting to note that even the acceleration due to gravity is not constant but reduces in value as you get further from Earth’s surface. Over the range of altitudes we are dealing with however, the change is negligible and so we can assume that it is a constant. Step 4 is to define the net force which is applied to the rocket during the thrust phase. This is where our calculations up till now, come together. Fnet = FT + FW + FD = 5 - 0.5 - (1.6 x 10-4) v2 Step 3 is to define an equation which will allow for aerodynamic drag. We need an equation because drag is not constant. The last two values that we’ve found can be assumed to be constant. With drag however, the value is dependent on the velocity of the rocket. In fact, it is dependent on the square of the velocity – if the velocity is doubled, then the drag increases by a factor of four. The equation that we define will form the last term in equation 2 which will lead us to our quadratic (equation 4). = 4.5 -(1.6 x 10-4) v2 To allow for aerodynamic drag, we’ll calculate the Force of drag: FD = ½ ρair ACS Cd v2 = (-1.6 x 10-4) v2 equation 1 Where: ρair = 1.225 kg/m3 (Which is the average density of air at sea level.) (For the purpose of these calculations, ρair is assumed to be a constant.) ACS = 4.8 x 10-4 m2 -- This is the cross-sectional area of the body tube at it’s maximum diameter. The Estes equation 2 It’s interesting to note that although the motor is producing 5N of thrust, only 4.5N of force goes towards accelerating the rocket. To put it a different way, the rocket will not go anywhere until the thrust is greater than the weight of the rocket. Also, when v = 0 the last term cancels out. Now that we’ve defined the net force, we can define the acceleration. The acceleration during the thrust phase is given by: a1 = Fnet /m (= Δv /Δt ) = [4.5 -(1.6 x 10-4)v2]/0.05 = 90 - (3.2 x 10-3)v2 equation 3 Now, Δv /Δt = (the change in velocity) / (the change in time). Normally, Δt is a very short time interval but we can apply it here. Since we started at v = 0 and t = 0, we will let Δv = the velocity at the end of the thrust duration (which is approxiContinued on page 5 Rocket Themed Christmas Ornaments • Adorable Ornaments with a Rocketry Theme • Hand-Crafted! Fun, Cheerful and Brightly Colored • They Remind Others of Your Rocketry Passion www.ApogeeRockets.com Page 4 ISSUE 323 OCTOBER 9, 2012 Continued from page 4 Calculating the Coast Delay Period We have found Vmax = 42.2 m/s mately equal to Vmax ) and let Δt = the thrust duration (t1). We need to transpose (re-arrange) the equation for acceleration: A couple more simple calculations and we will know the altitude at which the thrust period ends. The average acceleration of the rocket over the ½ second interval is found thus: 90 - (3.2 x 10-3) v2 = Δv/Δt a = Δv/Δt = 42.2/0.5 = 84.4m/s2 90 - (3.2 x 10-3)v2 = Δv/0.5 We can determine how far the rocket has travelled in that ½ second like this: Δv = 45 – (1.6 x 10-3)v2 We will let Δv = v and transpose to give: s1 = ½ a t2 = (½)(84.4)(0.5)2 = 10.5m (~34feet) equation 4 (1.6 x 10-3)v2 + v – 45 = 0 (about 8½ g’s) Step 5 involves using ‘the quadratic equation’ (for lack of a better name). The beauty of it is that for any quadratic (any equation that contains one variable for which two is the highest exponent), you can find a solution. This is known as finding the ‘roots’ of the equation. To summarise so far, the rocket has accelerated from zero to approximately 42m/s (140 ft/s) in about half a second and reached an altitude of 10.5m (35ft). If you can imagine a video of the rocket’s flight, where we’ve hit the ‘pause’ button at burnout we would see the rocket at about 10.5 metres above its launch point. Equation 4 is such an equation. That is, equation 4 is a quadratic of the general form: Calculating the Delay Period av2 + bv + c where: a = (1.6 x 10-3) b=1 c = -45 The rocket’s maximum velocity (42.2m/s) coincides (approximately) with motor burnout and the motor’s delay element is now burning. The question now is, for how long will the rocket coast before it reaches apogee? To simplify the remaining equations, we will assume that at the beginning of the coast phase t = 0. The acceleration due to gravity is constant and is: We may find v thus: ag = -9.8 m/s2 (= afinal ) v = [ -b +/- √(b2 – 4ac) ]/ 2a The initial acceleration due to drag will be: = [-1 +/- √(1 - 4 x (1.6 x 10-3) x -45)]/[2 x (1.6 x 10-3)] ad = (-3.2 x 10-3)(42.2)2 = [-1 +/- √(1 + 0.288)]/(3.2 x 10-3) = -5.7 m/s2 = [-1 +/- 1.135 ]/(3.2 x 10-3) We’ll now reject the negative case, giving: be: v = [-1 + 1.135 ]/(3.2 x 10-3) The net initial acceleration during the coast phase will Continued on page 6 • Big Kits with Classic Styling and Bold Graphics • All Rockets Feature Laser-Cut Plywood Fins and Rings NEW! • Easy-to-Build. Durable. Exciting, and a Real Joy to Fly! Sold Exclusively at ApogeeRockets.com ISSUE 323 OCTOBER 9, 2012 www.ApogeeRockets.com Mid & High Power Rocket Kits! Everything Rocketry North Coast Rocketry Page 5 Continued from page 5 Calculating the Coast Delay Period Note that when t = 0 equation 5 equals the net initial acceleration. This is why we ‘re-set the clock to zero’ at the beginning of the coast phase. ag + ad = -9.8 + -5.7 = -15.5 m/s2 (= ainitial ) Here is where the calculus comes in. If we have an equation for acceleration, we can obtain an equation for velocity by integrating the equation for acceleration. Don’t worry if you don’t know how to integrate an equation, that’s been done for you below and there are some notes at the end of the article on how to do it if you can’t see the pattern here. Note that I say initial acceleration and that it is negative. Both gravity and aerodynamic drag are slowing the rocket. As the rocket slows, drag will have less of an effect and gravity will eventually be the only force acting on the rocket. When the rocket falls and is accelerated towards the ground, the force exerted by aerodynamic drag will become positive but we are not concerned with that part of the flight. Hopefully the parachute will deploy and prevent our rocket from reaching a high enough speed to cause damage to either the rocket or anything on the ground. Integrating equation 5 will give the velocity with respect to time. v = -9.8t + 5.7 e-t + (42.2 – 5.7) = -9.8t + 5.7 e-t + 36.5 Using Euler’s Number In the title of this article I mentioned Euler’s number and you may be wondering now what it has to do with rockets. Euler’s number is represented by the symbol ‘e’ and is approximately equal to 2.72. It is a value that is useful in explaining growth and decay cycles in biology, sociology, economics and physics where the rate of change is proportional to the value. The nice thing about e is that when used in an equation, the graph that results is a smooth curve (a curve not unlike the acceleration curve which we see when a rocket is subject to a negative acceleration during the coast phase). By incorporating e in the equations below, we can approximate what is happening to the rocket as a result of gravity and aerodynamic drag. The acceleration with respect to time is given by: ac = afinal + (ainitial - afinal )e-t = -9.8 + (-15.5 + 9.8)e-t = -9.8 – 5.7e -t equation 5 equation 6 Note at a constant is added (in brackets) so that when t = 0, v = 42.2 . Integrating again gives the displacement (s) with respect to time. Again, a constant is added so that when t = 0, s = 0 s = -4.9t2 - 5.7 e-t + 36.5t + (5.7) = -4.9t2 + 36.5t - 5.7 e-t + 5.7 equation 7 Equation 7 resembles a quadratic. I say resembles because the 3rd term has an exponent of –t. As t increases, the 3rd term (5.7e-t ) will tend towards zero so we will ignore this term and treat equation 7 as though it is a quadratic. This is ‘fudging’ the results a little, but we are only concerned with an approximation so it is a fair assumption. The advantage of this is that we can apply the usual rules that apply to quadratics. Since the ‘a’ term in equation 7 is negative, s will have a maximum value when t = –b/2a. Continued on page 7 We’re Paying Cash For Great Articles for This Newsletter Are you a writer looking for some serious pocket change? We’re paying up to $350 for good how-to articles for this newsletter. If you’re interested, see our submission guidelines on the Apogee web site. www.ApogeeRockets.com/Newsletter/Newsletter_Guidelines Page 6 ISSUE 323 OCTOBER 9, 2012 Continued from page 6 Calculating the Coast Delay Period becomes:[-9.8/(1+1)]t(1+1) = -4.9t2 Term 2 of equation 6 is: 5.7 e-t The delay period will be approximately: To integrate an e term, neither e nor its exponent is changed. tdelay = -(36.5) / 2(-4.9) = 3.7 seconds Substituting 3.7s in equation 7 gives the altitude gained during the coast phase: s2 = -4.9(3.7)2 + 36.5(3.7) - 5.7 e-(3.7) + 5.7 The coefficient is divided by the exponent’s coefficient: (5.7)/(-1) e(-1)(t) = -5.7e-t About the Author = 73.5 m (241 feet) Recall that the altitude at the end of the thrust phase was: s1 = 10.5 m The altitude at apogee will be approximately: s1 + s2 = 10.5 + 73.5 = 84 m (275 feet) A crash course in integration If you’re struggling to see how equations 6 and 7 were obtained, here’s some more detail: Ian Voss is a member of the Queensland Rocketry Society (QRS - www.qldrocketry.com) in Australia. He enjoys attending the regular club launch days and designing and scratch-building rockets in his spare time. He is currently the holder of an altitude record (for an A8-5 motor) and looks forward to gaining his L1 certification in the near future. Ian firmly believes in the benefits of getting involved in the hobby of model rocketry. “As a tool to promote learning, rocketry is hard to beat. It’s exciting and fun and captures the imagination like few other things can.” Term 1 of equation 6 is: -9.8t Which could be written as: -9.8t1 (Any number to the power of one equals itself) When we integrate, we increase the exponent by 1 and divide the co-efficient by the new exponent. Term 1 Minimum Diameter Motor Retainers! Apogee is your one stop shop for your minimum diameter rockets projects! • Fly High • Fly Fast • Impress Your Friends! We Have: • Minimum Diameter Retainers • Motor Extenders • Threaded Forward Closures • Adapters for Cesaroni Cases www.ApogeeRockets.com ISSUE 323 Ian and his son at a fete display to promote the QRS and rocketry in general. The rockets in the background belong to Ari Piirainen (far left) who is the Public Relations man for the club. OCTOBER 9, 2012 Page 7