Recent content by WillemBouwer

CWatters: Yes as stated in A it is a double acting actuator... Let's say that the duration of the strokes is the same so yes thanks, now you made that seem pretty obvious, haha... Thanks Now the other thing I started to have trouble with, say I do not want to place a spring for fail safe...

Hi all I am trying to set up a report showing the air consumption rate of my cylinder actuators on different valves. I am stuck at the very end where I should calculate the air consumption rate. All calcs are in metric units. Okay as an example let's use the following: Valve: SKG200 with...

I woud do it that way, you have the right thinking... you can also use that time and get the total distance and divide that by 2...

If you have an equation in mathematics: ax^2 + bx +c = 0 Then how do we simplify to get the roots? do you remember the term that c = (b/2)^2 for it to be the same roots. (x+c_new)^2... Try doing this... if you get the new c put it into the equation but remember to subtract it again... Really...

Yes Bill Nye Tho, for you it is very easy, but the purpose of this forum is to help people understand what they deal with by letting them do the problem themself, we just aid them when they struggle... Try helping not just showing that you know the answer...

I think if I tell you that you would realize how easy it is. So you have an object working to the right with kinetic energy = 0.5*mv^2 your answer here is incorrect, do the calculation again... Now what is the total forces working against the vehicle? friction and drag, draw a FBD for your...

if you have anything above the line you can make the numerator 1 and right it next to it. Like 4/6 can be written as 4*(1/6) its the same. Just show me your calcs, and I'll help you.

what is the answer of 1/(4^2)? 0.0625 now check what is the answer for 4^(-2)? This should give you some direction... 1/(U^0.5) can then be written how?

just ad a pic of your problem and also your attemot at a solution and we can figure some solution out

Very easy once you have done the equations for both the masses...

Attempt at a solution is not just 17... haha... Show your equations that you used, so we can help you... Did you draw your FBD's for the 2 masses?

How did you come to the 27.66 m? show your calcs without the drag force and I'll help with implementing the drag force...

Just a thought, this is what I would do, but do not take this as an answer, weight for the real clever guys to come and give an opinion. Translate the speed of the shuttle to angular velocity. This will give a different speed at different radiu'... If you take the observers neck as its turning...

Yes that is correct, I just wanted to show you that the equation you have also works: with W1 = mad_1 and W2 = mad_2 with d_1 = ut +0.5at^2 = 0.5vt and with the acceleation being constant the time for the second period: v/t_1 = (5v-v)/t_2 so t_2 = 4t_1 hence d_2 = ut + 0.5at^2 d_2 =...

Lo.Lee.Ta: Okay let's assume the acceleration is uniform. You stated that W1 = Fd = mad what is the equation for W_2? remember the distance is not the same... What equation can you use to get the distance travelled? Linear equation? Also what is the time it takes for the second part if we say...