Recent content by rafehi

Have always struggled reducing a real world mechanical problem down to a simple model. Working on it... Just to clarify with a FBD of the external forces: http://img259.imageshack.us/img259/1201/84664287.jpg However, assuming that the column isn't going to buckle under the load (not...

Is it really that simple? The base of the robot is what's accelerating, not the camera (housing). I guess we can use the robot as the reference frame and take the base of the rod to be fixed, with the housing accelerating, however will modelling the dynamic system as static be a reasonable...

Hi all, I'm currently designing a housing for a camera and omnidirectional mirror for a robot. The housing is connected to the robot through a rod of about 25mm diameter. ...{...}... (housing) ...|... (rod) ...(___)... (robot base) I'm trying to compute the deflection in the rod when the robot...

Yeah, that's what I'm getting at. So I could just connect the output of the potentiometer directly to an op-amp (say, an inverting amplifier) and use that as the feed into the motor? Would that resolve the issue of the motor stalling and starting at different resistances?

Thanks for that, vk. Was doing some reading on the topic last night and was just about to leave to buy diodes, transistors and an inverter IC to make a PWM.

Not sure if I'm doing something wrong, but using a 3V battery, I get approx. 5rpm, with a current of 40mA. 1.8V: 0 rpm, 85mA 3V: 5 rpm, 40mA 6V: 12 rpm, 41mA 9V: 20 rpm, 50mA 12V: 30 rpm, 54mA Calculating the resistance required when using a 9V battery gives: (9V - 3V) / (0.04A) = 150...

Just messing around at the moment and trying to get my head around it all, hence why I'm using the potentiometer.

I'm trying to control the speed of a DC motor by using a potentiometer, but am unsure what value resistance I should use. Having tried 500k, 10k and 500ohms, the motor stalled almost as soon as I turn the dial (a bit more leeway with the 500ohm but not much)...

2A because each side has length L/2 - so Ak/(L/2) = 2Ak/L. Is the reasoning right?

Is 3.5A the correct answer? For a), by adding a voltage source across the terminals AB, you have a voltage source in parallel with the resistor. The other two resistors are in series with another, i.e. all current entering BC must exit through CA (and vice versa). Because the two...

There are two resistors in series, which are parallel with the 3rd resistor and voltage source in each case. You should easily be able to find the equivalent resistance of the circuit and thus the current knowing this. For a), I get 3.5A as the answer.

So basically treat it as a series question? Defining the temps as TL, TM and TR, (left, mid, right), Qin as the Q coming in from the left, Qmid as the heat through the middle, Qout as the heat from the right, a*Qvol as the fraction going into the left, a*Qvol as the fraction going...

Homework Statement There's a slab of a material with temperature T1 on the left and the T2 on the right. The thickness of the material is l with area A. In the centre, there is heat generation Qvol in the centre, which is a thin rod. Find the heat transfer Q through the material. Homework...

Now I get what you were trying to say! So those small holes on the side of the couplers are for set (flat) screws? Thanks for the help.

Trying to attach it to a wooden axle. While I managed to get a D-ish shaped hole, the shaft grinded it round when the load got to heavy. Decided to just drill a flat screw through the axle onto the flat end of the shaft to keep it in place and it seems to be working well enough.