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Apogee is important

Why is apogee important in model rockets? Apogee is pivotal in the timing of the deployment of the parachute which allows the rocket to find its way to the ground safely, without damaging the rocket or the devices inside it. Timely parachute deployment from apogee ensures that the parachute will not tear mid-flight, so the rocket will be easier to track by eye on its way to the ground; this way fewer rockets will be lost on their way down from a flight. Late or early parachute deployment results in net forces acting against the parachute (in addition to the force of gravity it already has to act against) in flight, causing a larger strain on the parachute and rocket frame. In relation to the rocket apogee, a one second difference in the deployment of a parachute translates roughly to a change in 32 feet/second in velocity. While it is possible to create rockets and even possibly parachutes that can withstand theses forces, it is ideal to minimize these stresses wherever possible. These extra forces can cause rips and tears to less hardy parachutes or damage to the rocket frame, causing the rocket to plummet downwards rather than parachuting slowly down. Without optimal parachute deployment, the recovery stage of the rocket’s flight will fail. Exploring rocket apogee detection methods, and in particular the use of the Kalman filter in rocket apogee detection, helps optimize performance while minimizing costs and disadvantages.

Current mechanical methods

Current methods used to deploy the parachute are completely mechanical and are not completely electronic. The most often used method as well as the most primitive method of deploying the parachute is to make use of a second ignition to push the components of the rocket out and towards the nosecone. The maker of the model rocket will usually drill a hole into a tube, and pack a certain amount of charge powder (the amount depending on the length and area of the hole). After the thrust and consequent burnout of the rocket engine, the last of the burnout sets a second ignition that blasts the charge up the body tube towards the nosecone. The force of the blast splits the rocket and sends the parachute components out into the air to catch wind resistance. The recovery wadding (Figure 1) must be placed in order to protect the parachute from the explosive charge. As seen in Figure 3, the whole system relies on the original ignition to relay through the electrical leads to the ignition charge.

Components of parachute ejection system

deployment system diagram
A diagram depicting an example of a non-electronic parachute deployment system. From: http://www.nakka-rocketry.net/parasys.html

This method, rather than rely on any sort of data, relies on the pre-computed estimates of apogee and when it happens. The length of the tube that sends the ignition charge can only be adjusted in steps of about two seconds. Those two seconds could end up being the difference between late deployment or early deployment, and on-time deployment. In addition to the large source of error introduced by the lack of precision of the apogee-predicting parachute fuse, it can be difficult to pre-calculate the timing of apogee. Without relying on any real-time rocket data, this non-electronic method of parachute deployment is can be unreliable. Accurate digital detection of apogee, is therefore key to an electronic method of parachute deployment.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Digital detection of rocket apogee. OpenStax CNX. Dec 18, 2013 Download for free at http://cnx.org/content/col11599/1.1
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