# Tag Info

0

Picking up from NMech's answer: For the step response, integration constants can be expressed in terms of $\zeta$ and $\omega_0$ According to swarthmore.edu Where $\omega_0 = \sqrt{\omega_1\omega_2}$ with $(\omega_1,\omega_2)$ being the two characteristic frequencies, and $K$ is the step amplitude After that, I believe $\zeta$ can be extracted from any two ...

1

The logarithmic decrement is not applicable to the overdamped systems because there are no oscillations. Also, there is not (at least to my knowledge) an equivalent to the logarithmic decrement approach to calculate the damping ratio $\zeta$ for overdamped systems. Free vibration measurement As AJN proposed what you can do is given the equation that ...

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 ...

0

Any parametric cad software can create a sketch and then measure the angle as driven. See onshape (browser based no need to install anything) SOLIDWORKS Siemens NX CATIAetc Also you can look at geogebra-geometry which is something that combines 2d algebra and geometry.

1

I found a related question that is effectively asking the same question. This is typically known as a four bar linkage (I only drew three in my diagram since the 4th bar is static). As far as I can see, you can derive a trigonometric relationship between the angles, but it isn't exactly simple.

0

1.) "... if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right? In a general sense, your thoughts are fine, but let's be more specific about the different scenarios: (1a.) - If the spring is stretched beyond its breaking strength, it breaks and would no longer work. (1b.) - ...

2

Just to be clear, there are external forces at work here! When you are extending the spring (or whatever), the box doesn't move downwards due to the restoring force of the ground it's sitting on. If you tried to make the box move while it was floating in zero-gee, conservation of momentum would make that impossible.

0

Question 1: If we don't pull the spring hard/far enough, when we release it it wouldn't provide much force. But if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right? You are right. The effect you are describing is called creep. most materials suffer from it. However, in this ...

2

Depending on how high you want it to go, you may not even need a spring, just a solenoid with sufficient travel and weight to it. The force generated will depend on $F=ma$ with the $m$ of the thing moving and how fast it starts and stops moving, but this is probably difficult to determine. It would be better to run some tests.

0

This is not an answer but to express how would I attack this problem. At any time $t$, there is a resistant force at the toe of the moving body, and the motion will continue if it is surpassed. Assume the sketch below is at $t = 0$, and $d = 0$, your job is then to find the energy accumulated between $t - t_f$ and $d - L$, in which $L$ is the length of the ...

Top 50 recent answers are included