11

I think you're overcomplicating things. To push 85 kg up a 15% slope against gravity of 9.8 m/s2 requires a force of $$ \sin (\arctan (0.15) ) = 0.1483 \approx 0.15 $$ $$F = 85 kg \cdot 9.8 \frac{m}{s^2} \cdot 0.15 = 125 N$$ With an 80 mm wheel, this requires a torque of $$T = 125 N \cdot 0.04 m = 5 Nm$$ To do this at a forward speed of 11.11 m/s ...


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 ...


6

Six bar linkages are pretty easy to find. Six bar is the go to guy when a four bar does not offer enough movement flexibility. Bigger European style hidden hinges are nearly all six bar mechanisms, essentially they are two four bar mechanisms that share part of the structure. These are common in many uses but the sixbar hinge you can probably find at home ...


5

I recommend researching motor driver circuits. These motor drivers internal has a H-Bridge design and have both direct on of digital signal to manage both direction (forward, backwards) and stop. You will also need a micro controller to control the motor driver. Below is an example block diagram for you to visualize. Below are few good places to start ...


5

I think it might be shorthand for a kinematically constrained design. It is the "This height gauge has three points of contact with the surface plate which makes it kinematic." which makes me suggest this interpretation. I also found a link where they seems to equate a "kinetic mount" to a mount that is "is fully constrained, but not ...


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

The angle $\beta$ can be obtained from the geometry $$ 0.2*1000 \sin (\beta )+50 \sin (60 {}^{\circ})=150 $$ This gives $\beta = 0.5627 rad$. Assuming the angular velocity of the arm relative to B is $w$, the velocity of the point D can be computed as $$ v_D=\frac{2000}{60} (2 \pi ) \{0,0,1\}\times \frac{50}{1000} \{-\sin (60 {}^{\circ}),\cos (60 {}^{...


3

Let’s consider gears as being represented by a pitch circle on which lie several equally spaced dots with a number equal to the number of teeth. These dots can either represent the teeth or the gaps between teeth, and they follow the rule that, as a pair of meshing gears rotate, the dots of one gear will coincide with the dots of the other gear as the dots ...


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

The velocity of P' is perpendicular to AD, and the velocity of P is perpendicular to BP. So clearly the directions are not the same. Thus there is relative velocity between P and P' as shown in the diagram. Now, if gyroscopes are mounted at D, P', or any point on AD, except A, they will all measure the same angular velocity. A gyroscope mounted at P will ...


3

I often approach this type of problem by identifying constraints. Each independent constraint limits the number of degrees of freedom, required to completely define the system. $$DOF = DOF_{system} - \sum \text{Contraints}$$ In 3D systems, each body has 6 degrees of freedom: Translations in $(x, y, z) \rightarrow 3 \;\text{DOF}$ Rotations $(\...


3

There is a fascinating historical collection of mechanisms compiled in the late 19th century here: Mechanical Movements and somebody is working through the task of making animations of them for the website. Manufacturing technology has progressed since then, but the basic ideas are timeless. We used to have a printed copy of this book in our "office library"...


3

Okay, it took me way too long to figure this out. I'm thinking this is just not a very intuitive interface. Here's the fix for anybody else who makes it here. Click #1, Click #2 (carefully!), then check the box at #3.


3

A simple machine is what you're referring to. A simple machine turns motion into mechanical advantage. Examples include: lever inclined plane pulleys wheel & axle gears inclined plane/wedge/screw These machines convert extra motion or rotation into mechanical advantage so that the same work can be done with less applied force. As for this: How can ...


3

In order to make the notation shorter for the remaining calculations in my answer I will denote the matrix with $$ H = \frac{1}{2} \begin{bmatrix} -q^\top \\ q_0\,I_3 + q^\times \end{bmatrix}. $$ In this case you have four equations but only three unknowns. So there might not be a solution that exactly matches all equations, especially when the time ...


3

Since considering four-bar linkages is a useful tool in my research field, I happen to know of a paper that precisely solves what you want to do. George H. Martin published in 1958 in the journal Machine Design the paper "Four-Bar Linkages", the following equations are taken from that paper: When you consider a four-bar linkage as shown below, then ...


2

Answer The other ground link is not shown and neither is the floating link. This diagram only shows The 4 desired output link positions The 6 poles generated from those positions The process involved to obtain 1 link Your first two points are correct, however, P23 and P'23 do not form the next link joint. P23 is the pole of the rotation from DC to CE and ...


2

a shock load on gearing is imposed when the gear train is forced to start or stop turning very suddenly, or when a heavy load is suddenly applied to a set of rotating gears. A good example of this is when the driver of a car with a manual transmission slams or "bangs" the gear changes without skilled use of the clutch, causing the car to suddenly lurch. This ...


2

Answering your first question: In the control field of mechanical engineering, a spring can indeed be a link. In the book Modern Control Engineering by Ogata, Katsuhiko the spring is frequently used as a link! Answering your second question: "Reuleaux called the ideal connections between components that form a machine, "kinematic pairs". He distinguished ...


2

When you constrain a beam end to point in the same direction as the clamped end (which is the case with the lower beams), then you have to apply a moment equal to $Ql$ (where $Q$ is the end load and $l$ is the length) to maintain this angle. You'll note that this results in an end-moment magnitude of $Ql$ at each end, which produces symmetric curvature. In ...


2

It is because you are using the trace of the matrix, which is equal to the sum of the singular values. The ideal $\mathbf{JJ^T}$ has singular values all equal to 1, but for an $n \times n$ matrix this sum would be equal to $n$, not $1$, and the text is defining the ideal matrix norm as 1.


2

Each car depending on its handling has a maximum safe turning speed $v_{max}$, and radius $r$. Let us say the current speed $v<v_{max}$, then your turning angular velocity is $$\omega= v/r.$$ We need to go an arc of $\pi/2$ so the time it takes to turn is $$t=\pi/2\omega= \pi r/2v$$ which will give the decision distance $x$ as $x=t\cdot v$. Edit Above was ...


2

No. You cannot increase both torque and speed without adding more energy to a system. In a gear train/belt & pulley combo, you have one member rotating slower but with higher torque and the other faster but with lower torque. You would need an outside power source to further increase the speed or torque beyond what you input, per the conservation of ...


2

There is no need to know the mass of the block. If it were movable yes, but in this case you can assume that is a fixed wall. So the whole idea is that you are assuming a constant (average) deceleration of the bullet. Since you already know the deceleration on the bullet, then from the equation $$\sum F = m\cdot a$$ you know that the only force on the X axis ...


2

Let's assume for now the wedge S massless. its mass is not going the change the aspect of the behavior of the One kg mass. The components of the 1 kg along and perpendicular to the ramp are $$F_{parallel}=mgsin37 =1*9.8*0.601 \quad and \quad F_{perp} =1*9.8cos37$$ The acceleration of the wedge S is $$\alpha =F_{parallel}/1=gsin37=9.8*0.601=5.89m/s^2$$ And ...


2

Frictionless surface. Assuming the initial velocity is perfectly tangential to the circular motion to begin with : greater than $u_0 >\sqrt{g R \sin\theta}$, then the car will spiral outwards less than $u_0 < \sqrt{g R \sin\theta}$, then the car will spiral inwards exactly equal to $u_0 =\sqrt{g R \sin\theta}$, then the car will continue to move on ...


1

In gear design for different machinery there is a combination of loads that causes damage and excessive wear and the equation for that load: F* = Ka Kv Km Ft Ka is application factor inherent in connecting to moving parts. Kv is dynamic factor caused by imbalance and deficiency in gears. Km is load distribution factor on gear. Ft is service design ...


1

I remembered the concept but not the name so I searched a few terms and found that Wikipedia calls it a quick return mechanism. Many such mechanisms are linkages (requiring no springs and catches, unlike your idea) and therefore may be called more specifically quick return linkages.


1

Conventions vary. As defined on that page, the deformation field $\boldsymbol{\chi}_\kappa$ represents the positions of the current points/elements/particles in an object right now as a function of a reference configuration (designated configuration $\kappa$) that typically corresponds to where the points were earlier: $\boldsymbol{x}=\boldsymbol{\chi}_\...


1

You have half the answer in your question. The device is known as a level wind mechanism and is a groove cut into the drum for the follower/rod in your drawing. A segment of video on the YouTube will show you a general shape for this mechanism. It's effectively a helix that reaches from end to end with a gentle curve to reverse direction (and cross the ...


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