# Tag Info

13

It is done carefully. Lots of other rotating machinery has the same problem, and whole systems exist just to deal with it. For example, jet engines are usually smaller, but also usually spin much faster. Balancing a jet engine is something that gets lots of attention at manufacturing, and again any time when it is put back together after having been ...

10

If we simplify the whole bridge into 2D thin beam with a constant section size, no internal damping and subject only to small vertical deflections, then the natural frequency is determined by simple harmonic motion: $$n_0 = \frac{1}{2 \pi} \sqrt{ \frac{ k } { m } }$$ Where $n_0$ is the natural frequency, $k$ is the ratio between restorative force and ...

10

Designing the lift system for a hovercraft isn't actually about the mass and the acceleration, it's about the pressure required to lift the hull and the rate at which the air leaks out from under the hull and needs to be replaced. Together, these determine the lift power required. Horizontal motion is a completely separate issue, which requires knowledge ...

7

Convert the nonlinear model to state-space form, $x'=f(x,u)$, and linearize it to get a linear state-space or transfer function representation. You can use this to design a PID controller and simulate it that together with the original nonlinear model to see how the controller performs. As noted, the performance of the linear controller will most likely ...

7

Flywheels are used in space, but not for storage of energy. They are used for attitude adjustment, as the spinning of a large mass has significant implications in an object which is free to move. Unlike on Earth, where a flywheel or gyro can be spun up against the force holding it in place (bolted to foundations), and change in spin speed will cause the ...

6

Let's first compute the model. The control design is a separate effort. The torque applied to the drum is $n T_M$, where n is the gear ratio and $T_M$ is the output produced by the motor. $T_M= K_T i(t)$, where $K_T$ is a proportionality constant and $i(t)$ is the motor current. Now we can write the equations for the mechanical system: $$m y''(t)+m g-k (... 5 The ESA has a page on compressor blades. They give a good dimensioned diagram of an approximate shape; here are some basic dimensions: Length: 300 mm Width: 30 mm Height: 70 mm Thickness: 5 mm I can't find any complete open source designs (i.e. high-quality, technical engineering drawings), but this is a good approximation. 5 There are two things going on. First, even if this "actuator" can produce constant torque, the torque required to keep the load spinning will be at least in part a function of the spinning speed. There will be some friction and other forces that increase with increased speed. Viscous friction increases linearly with speed, and other effects, like air ... 5 Take the Fourier Transform of the time varying driving force, this will give the frequency content of the driving force. Multiple modes of vibration can be driven at once, and will superpose with each other, but time varying driving forces with a frequency content that are high at frequencies near a particular resonant frequency will mainly drive the ... 5 To obtain the modes shapes and resonant frequencies, you start from your equation of motion with no externally applied forces, which is indeed as you've stated.$$\mathbf M \mathbf{\ddot q} + \mathbf K \mathbf q = \mathbf 0 \qquad (1)$$For brevity, I've let \mathbf K = \mathbf K_b + \mathbf K_m. Currently, \mathbf q(t) is a function of time. If the ... 5 This is quite easy: Choose one of the wheels build a second wheel so it would satisfy Ackerman condition Get the First rotation point ( O1 ) Do the same with the second wheel, and get a second rotation point ( O1' ) The actual rotation point can be anything between these two points. If you choose, for instance, the first point as rotation point, it means ... 5 You won't get much resonance because the phase the thing is being driven with keeps changing. The resonant thing will act like a notch filter, so you are left with the frequencies near its resonance. However, for resonant energy to build up, the system has to get pumped for a while. While a short segment of the filtered white noise could resonate the ... 5 You didn't account for the acceleration of m_1. Setting up a free body diagram on the weight shows:$$m_1g - T = m_1a_y$$Where T is the tension in the rope, a_y is the acceleration of the block. This leads to the following corrections:$$\tau_1=\mu(2\pi Rh)\frac{R\omega}{a}*R$$(The original had a value for force, whereas we need a torque.) ... 4 Advantages I'll start off by quoting from the "Benefits" section of the Wikipedia article: The load in a planetary gear train is shared among multiple planets, therefore torque capability is greatly increased. The more planets in the system, the greater the load ability and the higher the torque density. The more the merrier. The same idea is covered ... 4 It's a question of how many independent variables you need in order to write out the equations of motion. Let's walk through it step by step. Start by just considering Pulley 1 and Mass A. If they're joined by a massless inextensible cable (as shown in your figure) then the displacement of Mass A can be determined as a function of the rotation and radius ... 4 As I remember (learned it 15 yrs ago) You need to have an experimental data, which will help you to find "linear" sections and the their limits (with taking into account the things like hysteresis), so you will be able to find transfer function for each band. P.S. Why do you think that the servo motor system is nonlinear? Please describe the the mechanism ... 4 My question is how can I incorporate the axial force? Doing so would invalidate the assumptions made in the Bernoulli beam theorem, and therefore render your deflection equation invalid. Per the linked Wikipedia article (emphasis mine), Euler–Bernoulli beam theory (also known as engineer's beam theory or classical beam theory) is a simplification of the ... 4 With your code, and with c = 0.3, I get the following results: The amplitude and phase look OK. But the displacement does not show any damping. To see why read on below. I'm not sure about your notation. So the equations below may differ from those in your textbook. ODE The ODE you are trying to solve is$$ \ddot{u} + 2\xi\omega_{\text{res}}\dot{u}...

4

Stretch in the spring delta $Y = A.sin(\omega.t) = A.sin\sqrt(k/m) . t$ So the delta Y is not constant but if you are interested in delt Y_max delta $Y max = m/k$, by Hooks law. Because your system doesn't accelerate except at the beginning and end assuming the pulley starts and stops suddenly that's you maximum. Any gradual start/stop acceleration ...

4

The input force is 18.5+20.5 ft away from the fulcrum and the load is 18.5 away from the fulcrum. This means the force applied to the load is $\frac{(18.5+20.5)}{18.5} = 2.1$ times greater than the input force or for an input force of 50 lbs you get an output force of 105 lbs. However this assumes that all forces are applied perpendicular to the line ...

4

This is more of a long comment than an answer as i can not advice any specific software. First, if you intend to do anything professional in print or web productions than yous shouldn't be looking in direction of Microsoft for anything. Much abused does not mean any good. First tier would be to use direct vector drawing apps. In this category you have: ...

4

Large displacement shock absorbers used to dampen the effects of seismic events may be suitable for this application. These could be used to isolate the part of the structure that is grounded and the part that experiences the impact. Seismic dampers sit between the ground slab and the building and can effectively reduce the shock loadings in all directions....

4

The downwards tension force in a cable with a $130\text{kg}$ mass hanging under gravity is: $$F=mg=130*9.81=1275.3\text{N}$$ The upwards force provided by a servo with arm radius $0.02\text{m}$ and torque $13\text{Nm}$ is: $$F=\frac{T}{r}=\frac{13}{0.02}=650N$$ With two servos in your proposed system, the torque would be doubled, and so your theoretical ...

4

In addition to what Jonathan has said, there is several other practical things to consider: not only torque is limited, but power also; since power is torque times angular velocity (P=T*omega), your motor specification will limit the maximum speed you will be able to lift the load at depending on your application, you will not necessarily have to apply all ...

4

There is a slight mistake in your approach. It seems you have forgotten to add the contribution the moment due to inertial force acting at the center of mass $G$ to the moment equation. The moment equation should be : $$\sum M_A = -(15 m) N + (3 m) 140*10^3*a + (2.4 m) 140*10^3*g - (1.8 m) 140*10^3*a = 0$$ Evaluating for $N$ results in : $N = 2.57 * 10^5 N$ ...

4

Brakes primarily convert kinetic energy to heat energy. So a large area can absorb more heat lowering the peak temperatures ;of course this is strongly affected by the thickness/mass of the discs and other factors. AND the larger area can get rid or more heat . High temperatures cause deterioration of pad materials ,so lower ( not as high) temperatures ...

3

Perhaps it is helpful to compare the equation to its linear counterpart. \begin{align} \vec{F}&=\dot{\vec{p}}=m\vec{a}=m\dot{\vec{v}}\\ \vec{\tau}&=\dot{\vec{H}}=I\vec{\alpha}=I\dot{\vec{\omega}} \end{align} This linear formula says that the force on an object is equal the the rate of change of the linear momentum which is equal to the mass ...

3

Checking the units is an excellent way to double check your work; kudos for doing so. However, the next step in checking to see if your results make sense is to check limits. In your case, you can use physical intuition to identify how the system would act at very low frequencies, and how it would act without any damping. At very low frequencies ($s\... 3 Call the displacement of the box (where it is attached to the damper and spring) y(t). The force of the damper is $$-c\cdot \dot{y}$$ The force of the spring is $$k\cdot(x - y)$$ y(t) is the point at which the three forces on it balance to zero: $$F - c\cdot \dot{y} + k\cdot(x - y) = 0$$ x(t) depends only on the spring force and the mass:$\$k\cdot(x - ...

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