# Tag Info

1

Following on Spaderdabomb answer, the drag force $F_D$ acts on the body and, as such, it balances with the other forces and the inertia of the body ($\sum\vec{F} = m\tfrac{d\vec{u}}{dt}$). When you consider a situation where the body is at terminal velocity, this means its velocity is constant (at least in that direction) and its acceleration is null. ...

0

Adding to Peter Kämpf answer, these values for the drag coefficient pertain to flows where a turbulent wake exists in the lee side of the body, meaning that the drag is mainly due to pressure. For such flows the drag coefficient value does not vary with the Reynolds number. However, this is not true at low Reynolds numbers. For values below 1 inertial terms ...

0

The use of a Tesla high vacuum pump is indeed very compelling. I did some research during university for its use as a fluid pump, quite similar to the 2008 one you cited. I guess the design approach might be that of the many particles. At ambient temperature, the average speed of air molecules is about 500 m/s, which is comparable to the tip speed of a high ...

2

Draft and cavitation. Ships cannot afford to have big propeller diameters, they have to make do with the smallest diameter available in order to stay within the draft of the ships' hull. They operate in a medium which is 800 times denser than air, and one important concern is to avoid cavitation. This again means to limit suction peaks and leads to very ...

1

You are right to be skeptical of your answer. The technical specs show you should get something on the order of 100 m$^2$ . Your problem lies in basic trig: the force of the pressure is normal to the surface, and the surface is tilted by 5° - this means that the vertical component of force will be $\cos 5°$ times the normal force - you used the $\sin$ which ...

3

As I mentioned in my comments, there are many approaches you can take here, and full-on Navier-Stokes is not something you want to attempt. Either incompressible or compressible. Fully solving the Navier-Stokes is very, very far from real-time capable. Even for the most basic problems I can think of. You said you have incompressible figured out. That's ...

3

Hitting the air twice as fast takes four times as much energy ($E = {1\over 2} m v^2$). So there's an extra factor of two. To expand a little; you're fighting drag ($F_d$) in a rowing machine and drag increases as the square of velocity. ($F_d \propto v^2)$. Now, $Energy = force \times distance$ and $Power = {energy \over time}$ so \$Power = force \times ...

2

They can sometimes have this shape as they stretch downward from a faucet, as seen in this video, however, surface tension usually pulls the droplet into a roughly-spherical package. The reasons that raindrops have "streaky" trajectories in photographs is not that the raindrops are distended by air resistance, but rather that the short exposure time of the ...

2

The diagrammatic representation of rain is indeed incorrect. Raindrops do not resemble teardrops. This Youtube video by Minute Physics provides a really cool explaination of that: http://youtu.be/8lBvC7aFB40

1

Lift is approximately proportional to velocity squared of the aircraft, not the thrust. That is why runways are required. So the thrust is used the accelerate the aircraft to take-off velocity, which will produce enough lift to overcome gravity. Also, the fact that the thrust is less than the gravity in Antonov implies that it can't do this: ...

1

Multi-variable model-predictive process control technology has been around for some time now (at least since the mid 1980's). This means that the number of variables in the control problem is not an insurmountable problem. Assuming that the SpaceX process control engineers have a good intuitive understanding of the theoretical aspects of landing their ...

-1

Regarding lift more blades means more flying wing area which is a good thing up to the point where the following blades create too much turbulence. (2) blades is symmetrical balanced and efficient. Why would (4) blades not be , if not twice as good, at least significantly better and worth the cost to build a more complicated swashplate for collective ...

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I know this is an old thread. It was suggested that I post my answer to this same question from bicycles stackexchange. Link: http://bicycles.stackexchange.com/questions/10069/does-drafting-cause-resistance-to-the-lead-rider/32943#32943 Here is my previous answer: I recall this thread, and thought I'd add a link to a post describing an impromptu experiment ...

1

A rule of thumb exists if coriollis force is the dominant force balancing the pressure gradient. This is known as the geostrophic balance : $$\overrightarrow{V_g} = {\hat{k} \over f} \times \nabla_p \Phi$$ However if only a pressure gradient is being maintained by some source then the velocity will keep increasing as the pressure gradient results in ...

2

I have no direct experience with meteorology, but if you want the "rule of thumb", study the Euler equations. Specifically: $$\nabla p = - \rho\frac{\mathrm{D}\vec{v}}{\mathrm{D}t}$$ where D denotes the material derivative. That's the root of all other derivations.

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A propeller imparts momentum on the air. This creates a net velocity towards the rear of the propeller. Normally air molecules in the plane of the propeller have equal probability of moving forward or backwards; the propeller's action makes the situation asymmetrical. Since more air molecules move backwards, a low pressure area is created in front - and air ...

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