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The arrow ideally will fly out like a drag race car with the parachute deployed! A well designed arrow should have these properties. 1- Sharp and proportionally heavy point to accept a large momentum and deliver it as kinetic energy E= mv^2/2 2- long and balanced stem to accommodate a big arch and maintain separation between the tip and the fletching. 3- an ...


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In the movies, arrows shot into the air rotate so that during the descent, the arrow head hits ground first. What is the source of this angular momentum? An arrow shot on the Moon would not do that. Air and the geometry of the arrow are key. An arrow flying through the air is subject to two forces, gravity and aerodynamic drag. Gravitation will not make an ...


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Air. Conservation of angular momentum does infact dictate that whatever rotation it starts with it should end with, provided nothing else acts on it. Air allows its forward momentum to act on it. Consider a weather vane, a wind sock, or a flag. They rotate when not facing into the wind because one side presents more wind resistance than the other. Once ...


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As pointed out by dmckee in his comment, anyone (including myself) that has practised bow and arrow knows the arrow by weight and fletch inspection. As my English corrector points out, fletch doesn't seem to be a very commom word: it means that feather at the end of the dart/arrow. Everything resumes to how good is the fletching. When shot the arrow wants, ...


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The same reason objects which are heavier on one side tend to fall with the heavy side down: the tip of the arrow is denser than the rest of the arrow. The center of gravity is offset from its geometrical center, so the air drag, which is based on the object's geometry, causes a torque together with gravity as seen in this very professional picture of a body ...


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I have a Zuma 125 4 stroke gas motor scooter, and performed a lot of fuel efficiency tests during the last month. It had 8.98 hp @ 7500 rpms, and uses a CVT automatic transmission. I tested it on a backroad, with many different speeds and conditions. I drove 28.3 miles per test, starting with a full tank, and then measured how much fuel was used after the ...


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For fast moving objects like rockets the aerodynamic drag is dominated by inertial forces. Although air may seem weightless it actually weighs about $1.2$ kg per cubic metre (at ground level - the density decreases with altitude). A the rocket flies it has to push the air out of the way. That means it has to accelerate the air, and accelerating the air takes ...


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There is a similar concept used by ballistic missiles called a Drag Reducing Aerospike It is a flat disc deployed forward of the nose that... The aerospike creates a detached shock ahead of the body. Between the shock and the forebody a zone of recirculating flow occurs which acts like a more streamlined forebody profile, reducing the drag. ...


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A modern airfoil is designed on the basis of the desired pressure distribution over the chord length of both sides. In some cases, only a single angle of attack is relevant while in others the airfoil must be a compromise of the pressure distributions over a range of angles and flap deflections. Two parameters can be used to tailor the desired pressure ...


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Watch a video of a rocket launch, say the space shuttle. Within a minute of the flight the launch director might make a statement about maximum dynamic pressure. The launch vehicle is producing a supersonic shock wave and the interaction of the atmosphere is at its peak. Within another $30$ seconds to a minute the dynamic pressure with the atmosphere is ...


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When you're considering hydrodynamic forces there are always two things to consider. One is the viscosity of the fluid, i.e. how much energy it takes to make it flow, and the other is the inertia of the object, i.e. how much energy it takes to make it move out of your way. At high velocities the drag is dominated by the inertial forces. Air weighs more than ...


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Let's start with an airplane making a circle level to the ground at a consant speed. Its position is given by $$\vec{x}_1(t) = r\begin{bmatrix} \cos(\omega t) \\ \sin(\omega t) \\ 0 \end{bmatrix}$$ where $r$ is the radius of the turn, $\omega$ is the angular velocity (i.e., $v/r$), and $t$ is time. The $x$- and $y$-coordinates are parallel to the ground and $...


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There are lots of answers on this site regarding "how is lift produced" etc. but this post asks about the determination of a typical wing profile, so it might be appropriate to follow through on that aspect. Determine the chord (length from wing leading edge to the trailing edge). Decide on your camber line, for a flat bottom surface wing, this is the ...



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