# Tag Info

132

This answer is nothing more than a variation of Sklivv's answer. I simply wish to discuss some quantitative ideas following from Sklivv's answer and discuss what I understand (from an aerospace engineering friend) to be a common conceptual mistake - that the application of "mere surface effects" and "application of Bernoulli's principle" is wrong. These "...

91

You're talking about a device (in helicopters the tail fan imparting horizontal thrust) that counteracts the torque imparted on the main rotor (and therefore on the helicopter) by the surrounding air as the main rotor is dragged through the air. You propose instead to impart an opposite torque through a reaction wheel. That would indeed impart an opposite ...

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The vane has to be designed so that it has a preference to point in the right direction. In the example that you included, this is implemented by the flag at the back providing a broader cross section than the arrow head and also by the rooster standing slightly to the back half of the arrow. You are correct that if the vane became perfectly anti-aligned to ...

57

You are correct if your boat will only travel in a straight line. In real life the motion of the boat will often have a yaw angle, so that it is moving slightly "sideways" relative to the water. For example it is impossible to make a turn and avoid this situation. If the front is too sharp, the result will be that the flow can not "get round ...

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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 ...

52

When you would enter the water, you need to "get the water out of the way". Say you need to get 50 liters of water out of the way. In a very short time you need to move this water by a few centimeters. That means the water needs to be accelerated in this short time first, and accelerating 50 kg of matter with your own body in this very short time will deform ...

47

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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There is a YouTube video that visualizes the air flow around a propeller for various configurations. I caught a screen shot of a moment that more or less shows what is going on: As you can see, this happens at 2:07 into the clip - this happens to be for a dual rotor configuration (two counter rotating blades) but the principle is the same. Behind the rotor ...

29

Re-entry velocity from LEO is $~7,800 \frac m s$, from lunar space it is as high as $~11,000 \frac m s$ [1]. Different books give the terminal velocity of a skydiver as about $56 \frac m s$ or $75 \frac m s$ [2, 3]. The exact value isn't material, but the fact that it is two powers of ten smaller then re-entry velocity, is. The difference between skydiving ...

29

Any speculation about what shape might be best is meaningless without specifying the flow conditions. For the keel on a boat, the main one is the Reynolds Number, a parameter that is proportional to the the length multiplied by the speed. In most low-speed applications, a sharp leading-edge is not the best. With any incidence, the flow will tend to separate ...

27

Just a naive and very approximate calculation: If you neglect friction, google that human lung capacity is around $6\text{l}$, air density around $1\text{kg m}^{-3}$, take the inhalation time $6\text{s}$ and approximate that the nostrils have area around $1\text{cm}^2$, then the velocity of inhaled air will be $\frac{6\text{l}=6000\text{cm}^3}{6\text{s}\cdot ... 24 There are lots of questions here that I will try to answer, hopefully I'll get to them all... Creature Comforts It's hard to "just fly higher" when you consider passenger planes. Supersonic military aircraft like the SR-71 do fly ridiculously high. It's service ceiling is 85,000 feet! But, it has the advantage that it doesn't need to keep anybody but the ... 24 Let's look at this another way: you're just moving from one fluid to another. Sounds harmless, right? By specification of the problem, we're at terminal velocity when we hit the water. The force of drag (in both mediums) is roughly: $$F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A = \rho \left( \frac{1}{2} v^2 C_D A \right)$$ You can imagine that ... 24 Let's look at the relationship between momentum and energy. As you know, for a mass$m$kinetic energy is$\frac12mv^2$and momentum is$mv$- in other words energy is$\frac{p^2}{2m}$Now to counter the force of gravity we need to transfer momentum to the air:$F\Delta t = \Delta(mv)$The same momentum can be achieved with a large mass, low velocity as ... 23 It's not the falling that's fatal, it's the deceleration at the end that kills you. Something like water or concrete does this on a sub-meter distance (which requires extremely high forces). On the other hand a gas is much less dense, so it cannot decelerate a falling object nearly as quick. Sometimes inflatable cushions are used as safety nets (think: ... 23 I just can't wrap my head around why pressure decreases as velocity increases This is a classic misunderstanding of Bernoulli's equation. What Bernoulli's equation actually says is that the velocity will increase in the direction of decreasing pressure:$P_2-P_1=-\frac12\rho(v_2^2-v_1^2)$. This makes sense: if the pressure is higher on the left than on the ... 23 Being in orbit isn't about going very high, it's about going sideways very fast. In order to get to orbit, about 80% of the energy is used in achieving orbital speed, which only 20% is used in getting to orbital altitude. A re-entering craft needs to rid itself of its kinetic energy, but you can see that most of that energy actually doesn't come from its ... 22 Consider jumping into a swimming pool. Do a barrel-roll (sorry I mean cannon ball, that just kind of slipped out). It's fun, you enter the water nicely and make a huge splash, probably soaking your sister in the process (that'll learn her). Now do a belly flop. Not as fun. You displace exactly the same amount of water in the same time, but this time there is ... 19 I doubt if anyone has come up with a complete explanation, but some laboratory simulations have created similar patterns. They happen if the central and surrounding areas in a flat, circular disk of fluid have different velocities. Emily Lakdawalla at The Planetary Society covers it at this site. She also explains how other patterns (triangles & ... 19 I disagree with the most voted answer, by CAGT. He says "This area is completely different to the one above", but this means nothing. The equation$p = {F \over A}$mentioned by the author does hold, and there is no contradiction or paradox in it. In fact, the equation$p = {F \over A}$holds not only here but anywhere else in physics. You may write it in ... 18 Each window represents a restriction to the air flow. The greater the pressure difference across the aperture, the greater the flow. An electrical analogy: each window is a resistor. The current through the resistor is proportional to the voltage across it - but when you have two resistors in series they must carry the same current (air that enters through ... 18 I often wondered about these things - then I came up with a simple experiment that works for me because I have a simple bike computer (thing with a magnetic pickup on the spokes that updates my speed every second). I find a flat piece of road, and ride at a certain speed (say 20 mph on my road bike, or 15 mph on my mountain bike). I then stop pedaling at a ... 18 How does they move slowly in air, without falling down? One possibility is soaring using a ridge lift - typically a situation when the wind is approx. perpendicular to a mountain ridge. The air is lifted at the front side of the ridge and an eagle can soar in the lifting air stream. This can also work without the wind, Which is a situation of thermal ... 18 The key insight here is that you have a lot of energy, and all that energy has to be turned into heat. There's no other way to lose it. Your goal is 0 m/s at 0 km height. It doesn't really matter what the original proportion of kinetic versus potential energy is, both need to be turned into heat, and that heat needs to be shed. Heat shields are very ... 17 The distances and speeds involved are materially different. On the scale of a parachute dive, the atmospheric density doesn't change much (and is relatively high). A parachutist quickly reaches a terminal velocity where the drag from the air matches the pull of gravity. In a re-entry, you're approaching in a much less dense atmosphere, and you're going ... 16 The books you have mentioned, are written from an engineering point of view rather than physicist's. The following books are recommended: Fluid dynamics For Physicists by T. E. Faber (Introductory. Doesn't need more backgrounds than you mentioned) An Introduction to Fluid Dynamics by G. K. Batchelor (Introductory. Doesn't need more backgrounds than you ... 16 According to Sighard Hoerner's Fluid Dynamic Drag, this would be the half-sphere with the open side exposed to the wind. Its drag coefficient is 1.42. A rod with a hemispherical cross section will even have a drag coefficient of 2.3 (right column in the graph below). If you restrict the competition to solid objects, still the half sphere wins with a drag ... 15 This is a very good question! Drag due to viscous effects can be broken down into 2 components: $$D = D_f + D_p$$ where$D$is the total drag due to viscous effects,$D_f$is the drag due to skin friction, and$D_p\$ is the drag due to separation (pressure drag). The equation above demonstrates one of the classic compromises of aerodynamics. As you mention,...

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