# Tag Info

16

The testing of ventilation systems and of building air-tightness is done using smoke sticks or similar - non-toxic smoke generators that produce no ash residue. One of these may be suitable for your needs. Smoke generators typically use mixes of two or more of: alcohol glycol glycerol water I suppose strictly speaking they're fog generators rather ...

14

It is preferable for the wing root to stall first. If the wingtip stalls before the root, the disrupted airflow near the wingtip can reduce aileron effectiveness to such a extent that it may be impossible to control the airplane about its longitudinal axis. When an aircraft stalls at the root first, it means there’s enough airflow over the tips of your wings ...

12

I won't say it's not possible to make a noticeable difference. But I would say that it's pretty darn unlikely. The lift and drag forces on any body generally depend on the velocity of the body ($v$), the density of the fluid ($\rho$), the area of the object ($A$) and a dimensionless coefficient ($C_L$ or $C_D$). The applicable area can be a bit fuzzy in ...

11

Believe it or not, we could have done this 50 years ago, if government funding hadn't been pulled from a project at the last minute. Frustratingly, after years of work by scientists, engineers and technicians, the Boeing X-20 Dyna-Soar project was cancelled just after work had started on the actual spacecraft. Here's an artist's impression of the X-20: The ...

10

Cross-interference between aircraft is a high unlikely event because all commercial aircraft designs have to pass DO-160 environmental testing requirements. Among the DO-160 testing specification is EMI/EMC Testing. These tests include Radiated Emission, Interference and Immunity Testing. Part of these tests are to answer "Two aircraft are very close ...

10

Quadratically. NASA page #1 says that, as an approximation, $$\text{Lift}\propto v^2$$ where $v$ is the airspeed. NASA page #2 says that, for a better calculation, $$\text{Lift}=av^2+bv+c$$ where $a$, $b$, and $c$ are constants. This is odd because it implies that lift is not merely proportional to $v^2$, because there is the $bv$ term in there, as well as ...

9

Now, I wonder if an air-launched spacecraft in the next 20 years realistically could really fly us to the moon - that is, could they reach escape velocities? Air launch to LEO: Done now Air launch to lunar orbit - yes, but at 20%-25% of LEO payload Air launch to Moon and back to LEO: Yes, but with about 5% of LEO payload It is easy to overlook some ...

9

You don't want to decrease the downward force on the wheels because that is your primary means of slowing down in an emergency. It already takes a while to stop a train, don't make it any longer. Remember that kinetic friction (wheels locked) is proportional to the downwards force on the wheel and the brakes are spread along to each car. Train wheels have ...

9

To add to what Russell said, it would take a very stupidly designed protocol to allow the comands from one aircraft to control the other, even assuming that cross-reception is good. As a everyday example, think of a bunch of people talking on cell phones standing near each other. One person accidentally receiving another's conversation just doesn't happen. ...

9

You're not that far off. In your analysis, you've just neglected the fact that the gas will accelerate as it passes through the converging nozzle. This is basically converting potential energy, in the form of pressure, into kinetic energy, velocity. As a result, the gas will no longer be at 400 psi when it reaches the throat. The compact form to compute ...

9

Since orbiting isn't about altitude, but is instead about velocity (orbiting is the art of throwing yourself at the ground, and missing because you're going sideways so fast), the final goal is to have a heap of lateral velocity so that gravity is merely curving your course, instead of actually pulling you to the ground. The most efficient flight to orbit is ...

9

The function of the wind turbine is to extract as much as possible energy from the moving mass of air. The energy of a mass $m$ of air is $$KE = \frac{1}{2}m v^2$$ and the power is (because the mass rate is $\dot{m} = \rho A v)$ $$P = \frac{1}{2}\dot{m} v^2=\frac{1}{2} \rho A v^3$$ So, the energy in the air is a function of the velocity. As energy is ...

8

Modern communications systems are able to achieve essentially any desired data rate and degree of message integrity in any well-enough-defined environment. It's "just a matter" of the degree of effort, complexity and $that are needed. Designing a system to meet any desired data rate, number of users, and integrity is thus "just a matter of engineering". ... 8 As has already been mentioned, the primary relationship is that lift goes with the square of the airspeed. To give you some intuition as to why this is, consider what a wing does. As it moves along, it deflects air downwards. Lift is the upwards force from imparting downward momentum on the air the wing passes thru. Momentum is mass times velocity, and ... 8 It might be written that way because the transfer function of a second order system is written as: $$h(s) = \frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}$$ where$\zeta$is the relative damping ratio and$\omega_nis the natural frequency. So in your system \begin{align} \zeta &= 0.7 \\ \omega_n &= 52 \end{align} It's easy to see ... 7 A space engineer deals with objects in space: from satellites, space stations, space vehicles and space probes to space debris. Space means high launch vibrations, vacuum, radiation concerns, high temperature gradients and extremes, complex trajectories, next to no maintenance (meaning: built in redundancy)... An aeronautical engineer deals with objects in ... 7 Are any designs based solely on data from trial and error used in critical mainstream engineering? Usually not. And the reason is that trial and error is expensive and time consuming. As engineers, we are always working on projects with a budget and a deadline. Take your rocket example. Rockets are expensive. For sake of argument, let's just say it's1 ...

7

In a wing the normal situation is that the aerodynamic force is upwards (resisting gravity). You are right that there is some shear and a fair amount of torsion, but the result is that: the top side of the wing is in compression, while the bottom is in tension. So in essence what happens is the opposite from the picture below (the force is applied upwards, ...

6

In the Fluid Dynamics community about 40 years ago, the group was primarily divided into experimentalists and theorists. However, at that time CFD was quite new, had to be run on expensive supercomputers, and untrusted. It was quite common that a theorist or experimentalist would at best discount the results of the CFD, while others may totally disregard ...

6

In automotive testing we use baby oil. When heated it produces a thick white smoke that is non toxic in low volumes and doesn't stick to surfaces. I built my own smoke tester using a metal pressure fed wax-oil/underseal sprayer, I drilled a hole and inserted an automotive glow plug to heat the oil. You can then experiment with a pressure that works best for ...

6

The acceleration is given. However, the upward force is $6mg = F_{up} - mg$ $F_{up} = 7\cdot mg$ In case of no acceleration: $0 = F_{up,normally} - mg ==> F_{up,normally} = mg$ If we neglect the gravitational force experienced every day by this man, then we get $6mg = F_{up}$

6

why is it logical to have the position of shear center at the location of intersection of line of action of shear force and x axis? That statement isn't logical. I think you have misunderstood how the shear centre is defined. The shear centre is the point such that an applied force passing through the S.C. does not cause any rotation of the section. In ...

6

In general, most airfoils only approximately display the $2\pi$ lift slope as predicted by thin airfoil theory. That is because airfoils are not actually infinitely thin in practice, and will deviate from thin airfoil theory by a small amount. $a_0$ is left in the above equation so one can predict the elliptical finite wing lift slope (effects of aspect ...

6

The short answer is these terms are defined by standards organizations based on specific test methods. These methods specify the engine condition, environmental factors, fuel chemistry, inlet and exhaust systems configurations, and the correction factors applied to the measurements. The specific tests and standards used vary by country and by industry. So ...

6

There are propellers with different aerofoil sections. For example, the propeller attached to the Rolls Royce engine kept in my graduate college had symmetric aerofoil sections roughly about 25 % and smoothly varying to bottom flat airfoil sections at the rest. The aerofoil profile selections are based on the performance requirement, structural rigidity, ...

6

Knowing the derivation is important because it usually tells you what initial assumptions were made in the derivation and what the limits of applicability of the resulting equation are. Understanding both of these things is an essential skill all engineers must master.

6

The rocket equation says that every bit of mass you carry on your rocket is very expensive, especially if that mass is not fuel. Designers of rockets are willing to spend lots of time and effort making sure that their rockets are as light as possible, hence "mass is at a premium." This inevitably means that the tanks are as thin as is feasible. Stresses are ...

6

It depends on wing geometry. You need to look at the aerodynamic center of the wing, which is the center of effort for perturbation forces. The center of lift (unstalled) will usually be well forward of the aerodynamic center. As Cl increases, it is desirable to have the center of lift shift aft. This produces speed stability by lowering the nose, thus ...

5

I'm going to answer my own question because just before posting it I found an answer. The Otto Lilienthal Museum has a comprehensive list of Lilienthal's designs. One is listed as the "small wing-flapping machine." It didn't use a propeller (or jet engine, of course!) but instead used a small engine weighing about 22 pounds when fully fueled. Its wingspan ...

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