13

radian is a derived unit, defined as the ratio of arc length to radius. As the ratio of two lengths it is dimensionless.


12

I don't know how this function was obtained (or what it is supposed to represent). But you could try searching the literature to see if someone has published equations of the curves. However, looking at it I would suspect that there is not a straightforward expression for these curves. The best option I can think of is to use graph digitization software to ...


8

Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis. Its moment of inertia is $r d\theta dr (r \sin (\theta ))^2$. The moment of inertia about the $x$ axis of the complete sector: $$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{...


7

All EM radiation from a point source falls off in intensity with the square of the distance. LEDs can't magically change this basic physics. LEDs can be quite directional in their emission pattern. One possibility is that the orientation of the LED relative to the receiver wasn't exactly the same in all your measurements. Another is that you were fairly ...


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's $1 ...


7

for $y = f(x_1, ..., x_n)$, the sum of the partial differentials with respect to all of the independent variables is the total differential: $$ dy = \frac{\partial y}{\partial x_1}dx_1+...+\frac{\partial y}{\partial x_n}dx_n $$ For our case: $$ \dot{m} = \rho Av$$ $$ d\dot{m} = \frac{\partial \dot{m}}{\partial \rho}d\rho + \frac{\partial \dot{m}}{\partial A}...


6

I think the diverse range of names for the second argument arises from the fact that the convolution operation is so useful in so many different fields. It is helpful to recall what the convolution operation does before addressing the specific terms. Quoting from Wolfram Mathworld, "a convolution is an integral that expresses the amount of overlap of one ...


6

Kinematics and dynamics Those are the steps to solve problems of this nature. Analize the kinematics of the system. $\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $_{o}\vec{r}_{RP}$ $\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $R(\varphi) _{B}\vec{r}_{RP}$ $\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $\big(x_{k}î + 0j + 0k \big)$ + $\big(...


5

First off, always remember that garbage in = garbage out; so if your data is garbage then your statistics will be garbage. In this situation your optimal data would be something like Run Hours Until Failure and your entire dataset would have failed already. With this in mind you may want to choose a conservative number from whatever statistic you calculate....


5

My guess is that you probably need another differential equation for the angular movement, that will involve the inertia, such as: $$m_G l^2 \ddot{\varphi} = m_G g l \sin(\varphi)$$ which yields: $$ \ddot{\varphi} = \frac{g}{l} \sin(\varphi)$$ You can then maybe use the small angles approximation: $$ \sin(\varphi) \simeq \varphi$$ Check out the ...


5

I thought I would expand a little on the answer offered by Karlo. Long story short, I would not try to calculate the analytical time response of a system to a square wave. That would be a serious pure-maths exercise, and not necessary for most engineering applications. Instead, I would suggest using the step response analytical vs. simulated to validate ...


5

No. First, there is much much more to most engineering disciplines than analyzing the response of something. Second, even when that is part of the task, it isn't done by "adding oscillators", whatever that even means.


5

If you want to evaluate a continues time transfer function at a specific frequency $\omega$ in rad/s you substitute $s$ with $j\,\omega$. For a discrete time transfer function you substitute $z$ with $e^{T_e\,j\,\omega}$. In order to see why you have to substitute $z$ with $e^{T_e\,j\,\omega}$ you can consider the transfer function $z^{-1}$, which is a ...


5

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.


5

Should I, as an engineer-in-training hoping to complete research, focus on trying to understand equations to my satisfaction, or should I instead just become well acquainted the equations, their use cases, and their general behavior? Engineers apply equations to predict behavior. The critical requirement is to know the constraints of the equation. The ...


4

Think for a moment about what it means for $\eta(t)$ to have reached its steady state. It means precisely that $\dot{\eta}(t)=0$. If you plug that into your first equation and solve for $\eta(t)$ you get your solution as you've already observed.


4

If you don't have hard data, making assumptions (preferably "reasonable" ones) is the only option you have. (Maybe that's why engineers used to call their slide rules "guessing sticks...") You can't ignore the fact that most of the units have not failed so far. A plausible approach to this would be to use the times-to-failure that you know, to fit the ...


4

For the Maths of calculating uncertainty the standard document is the GUM. Which describes all the maths but can be somewhat unclear if you don't already have some idea how it is supposed to work. Depending on your current level of expertise there are several good introductions to uncertainty calculations. I would recommend A Beginner's Guide to Uncertainty ...


4

I think you are neglecting the broad range of applications included in solving 'some kind of differential equation.' At the very basic level of physics, everything we know about the world is expressed in terms of differential equations. Some examples: Electromagnetism (gives rise to most topics in electrical engineering) Mechanics (gives rise to most ...


4

You have a single-valued function dependent of two variables. There are many ways to model that. If you know something about what this function represents, then going back to the physics might yield a useable equation with only a few coefficients. Digitize a bunch of points, throw them at a least-squares error minimizer and see how close the result is. ...


4

Yes. The first derivative of the deflection is equal to the tangent of the deflection, which for small deflections can be approximated as equal to the angle of rotation of the beam at each point. The second derivative (times $EI$) is the bending moment along the beam. The third derivative (times $EI$) is the shear force along the beam. The fourth ...


4

Suppose Frame 1 is a world coordinate frame and Frame 2 local robot frame. The thing that's a little confusing here is that when you transform from Frame 1 to Frame 2, you are saying that you want to... transform frame 1 to frame 2. You want to take all of frame 1's points, and put them in frame 2. You want measure where points in frame 1 are located, but ...


4

For a detailed explanation search WikiPedia for derivative, partial derivative and total derivative. For a short, non-mathematical, summary see below. The partial derivative of a function of several variables is it's derivative with respect to one of those variables, assuming that all other variables are constant. The total derivative does not make this ...


3

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer. To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. ...


3

I think you are misinterpreting what your variables are. The focal length is fixed for a given lens and cannot be changed. What you can change is the image to lens distance. Combined these will determine what point is in focus using the lens equation. The data sheet for the camera you give, gives the focal distance as 6.68 mm. plugging this into $$\frac{1}{...


3

If you plot the x values of OP and HP, they cross zero, which seems unphysical. Looking more closely, you are doing your rotations backwards, i.e. you are doing our rotations in +theta whereas you want to go in -theta. This is something I often do too. To fix, just replace theta with -theta in M. Here's a figure of F when you do this which looks much more ...


3

The basic partial differential equations are relied on in many ways. If I have something already in my book of equations (Roark's Formula's), then I use that book for solving of the plate. These results were derived from the original equation and put in a more useful format If I don't have it in my book (for example, recently I had to find the equation for ...


3

$F=\frac{dp}{dt}$ is, in some sense, a more fundamental expression of Newton's law than $F=ma$ because $F=ma$ doesn't allow for situations with changing mass. You can easily derive the second from from the former by using $p=mv$ which gives $F=m\frac{dv}{dt}=ma$. Your question seems to be more about how to derive the expressions for the force at the ...


3

I don't have a simple formula but that is the way i would handle it. Assuming the four corners are circular with equal radius r, so that symmetry exists, the first moment of area of one half of the section should be: $$S_{half} = S_{half\,rect}-2S_{corner}$$ where $S_{half\,rect} = b(h/2)(h/4) = \frac{bh^2}{8} $ the 1st moment of area of the half ...


3

My calculation got the same answer as you. Here's what I did. You mentioned that the arm has some weight so I added that force to your diagram. Also, I labeled the center of the elbow joint as point 'E'. Assumptions The pully is mounted on a frictionless pivot (friction will reduce the amount of force required) The arm is perfectly rigid (if the arm ...


Only top voted, non community-wiki answers of a minimum length are eligible