# Tag Info

1

It looks like Pythagoras, but it is only remotely related. The important concept, as presented in SteveB's answer, is that the variables are considered to be independent, i.e. one does not affect the other. In mathematics, independent parameters are said to be orthogonal , and can thus be assigned to separate axes in Cartesian N-space. It just so happens ...

1

The general formula for error propagation is: $$\Delta f(x_1,x_2,\ldots)=\sqrt{(\frac{\partial f}{\partial x_1}\Delta x_1)^2 + (\frac{\partial f}{\partial x_2}\Delta x_2)^2 + \cdots}$$ Where does this come from? We assume that the errors are relatively small (ignore $\Delta x_i \Delta x_j$ terms etc.), and that the errors are independent (in the ...

1

The formula $$\frac{\Delta{A}}{A} \approx \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y}$$ is an approximation because you are ignoring $\Delta X$$\Delta Y A better approximation would be$$\Delta A=\frac{\partial A}{\partial X}\Delta X+\frac{\partial A}{\partial Y}\Delta Y$$Since errors always add we take the absolute magnitude of \frac{\partial ... 0 The square root is there as a better estimator of the error than just adding the errors together. If you add the errors together you are finding the maximum possible error which will happen when both quantities are a maximum(or minimum) together. This is an unlikely event compared with all the other domination of errors. The square root formula you you ... 1 Can an angle be defined as a vector? It depends on what you mean by "vector". If by "vector" you only mean something that has a magnitude and a direction, then yes, the axis-angle representation qualifies as a "vector". To a mathematician, a vector is something that is a member of a vector space. In this context, the axis-angle representation fails to ... 1 Describing a rotation as a vector, with the direction of the vector along the axis of rotation, and the magnitude of the vector as the angle, is known as the axis–angle representation. 1 If this article is to be believed, you would have no problem at all - in fact you could feel where the cities are, let alone the mountains. 1 The concept you're after is the dot product between 2 vectors (your displacements). More specifically you want to use$$ \vec{a} \cdot \vec{b} = |a| |b| \cos(\theta)$$to find$\theta$. 1 Here is another way to look at this problem: we should consider the total torque relative to the two points where wheels touch the ground. Clearly in your picture you rotate the whole bicycle backwards regarding the back as well as front wheel (points where wheels touch the ground). Imagine that the wheels are nailed. If you would pull the rope forward the ... 1 In the reference frame fixed to the bike, a backward movement of the pedal by$dx$moves the ground backward by$\alpha dx$. So the pedal moved backward$(1-\alpha)dx$with respect to the ground. If$\alpha>1$then this means it either moved forward, which would deliver energy to the string puller and thus makes no sense didn't move at all ($dx=0$) ... 1 I think the way to look at the problem is this: The tricky thing about this problem is that the force$F$is doing two things. First, it is pulling the entire bicycle backwards and, secondly, it is exerting a torque on the pedals. So let's try to separate those two effects. Consider, then, a person just sitting normally on the bicycle with his feet on the ... -1 It will definitely move backwards. The angular velocity of the wheel, multiplied by the radius of the wheel is much bigger then the angular velocity of the pedals, multiplied by their length (the ratio of angular velocities is constrained) so the backward motion of the bicycle is much more significant then the rotation of the pedals. When the horizontal ... 1 The ideal coordinates for this sort of problem would probably be these two red-dashed axes corresponding to the two actually-fixed points in your system. From this perspective, the center of the circle is at$(0, 0)$, the center of the ramp is at a point$(D, 0)$, the point circled describes a simple position$(a \cos \theta, a \sin\theta)\$ at a distance ...

Top 50 recent answers are included