The application requires that a 1000 lb horizontal load be moved in 1 second a distance of 30 inches. The cylinder size is 1-1/2" bore, by 1 " rod. The bulk modulus to be used is 200,000 psi. Tube size used were 3/4" OD x 0.065" thickness. Length of tube on blind end is 46.5". The length of tube on rod end is 38.75". I will give the frictional force as 580 lbs, because I cant understand how it originates... Ff= uN?!

Find the following:

1. acceleration time and acceleration stabilizing time
2. maximum velocity and maximum acceleration
3. Blind end and rod end acceleration pressures
4. flow rate required in blind and rod end

This is an example from a book and I would like to compare the difference in result to any other highly mathematical calculations.

The formula involve were like a plain algebra.

There is a reason why we need to include the tube length of blind and rod and it is very relevant for recalculation...

I can see "IMO" that all relevant data are given to solve this.

"The classical approximation of the force of friction known as Coulomb friction (named after Charles-Augustin de Coulomb) is expressed as Ff = μN, where μ is the coefficient of friction, N is the force normal to the contact surface, and Ff is the force exerted by friction. This force is exerted in the direction opposite the object's motion."

This simple (although incomplete) representation of friction is adequate for the analysis of many physical systems.


http://engineering.wikia.com/wiki/Friction

Regards Woody

This message has been edited. Last edited by: Woodygb,

Yes, woody, but is the .58 coefficient an assumed/estimate value? That's why I just give the frictional force because I don't know the reason why such coefficient just popped out in the calculation process. Maybe average for such size of cylinder and the corresponding load.

I believe ..but I may well be in error... and I was so an edit.

It seems that the load is being pushed along a surface with a friction coefficient of 0.58

Thus your 1000lb weight needs a force of 580lb's to slide it along.

http://www.ac.wwu.edu/~vawter/...s/Forces/Normal.html

This message has been edited. Last edited by: Woodygb,

Ok, anyhow with regards to tube/valve coefficients of flow, the way I understand about the inclusion of tubes is just to illustrate the difference of natural frequency as compared to when the valve choosen mounted directly or as close as possible to the cylinder.

This discussion does not involve any valve at all. There is no need of any pressure value since we are given the cylinder size and load.

I will try to answer to the best of my limited ability as possible for any clarifications, but as I said I have not designed any fast acceleration system(it requires natural frequency consideration). For slow moving systems, the natural frequency calculations maybe sustituted by estimate.

This message has been edited. Last edited by: maglub,

Peter, the book used x6, and 1/3 of natural frequency. Let's try to use this value so that the possible margin of error between your calculation and my book's will be reduced.

Woody, if you are interested, try using the bulk modulus formula and relate this to hooke's. The relationship of both can be applied to the natural frequency of the system. Once you get the possible natural frequency of the system, it will be smooth sailing for you...


Gudnite to all... Back to RoN .

1. You can't calculate the natural frequency until you know the volume of oil that is going to be compressed. You provide numbers for the tubing to the cylinder, but you don't provide the length of the cylinder. Just because the actuator moves 30 inches doesn't mean it is 30 inches long. The cylinder can be longer and the natural frequency lower.

Ok, we can just assume to be 30 inches and the actuator will stop there for simpler calculations. Otherwise if we add more stroke it will just reflect on the volume of the rod side where the natural frequency result will change due to the total rod piston face to rod end tube length volume. This is just an example and we can assume that a fixed infinite resistance is at 30 inches.

This example IMO is also a one way motion extending to calculate the unknown. The resultant flow to extend is more of relevance than the flow that exhaust from the rod size since for a differential cylinder flow in and out, the flow rate exhausting from the rod is smaller, almost a little more than half of required floe rate to blind in a 2:1 ratio.

Although consideration of the retracting flow must be taken, but will depend whether a slower retracting speed is fine(within the valve pressure drop/max flow limit) so as to utilize a better size of valve(economic).

IMO, maybe the reason why the natural frequency is 1/3 of the calculated NF because in actual application, there are other capacitances on the system which are not included in the calculation. So the useable acceleration is better estimated by dividing the calculated NF by 3.

This message has been edited. Last edited by: maglub,

The question should I think read like this...
Requirements Maximum speed and shortest time.

quote:

The application requires that a 1000 lb horizontal load be moved accelerated to a constant velocity over in 1 second a distance of 30 inches in 1 second and then decelerate to a stop. The cylinder size is 1-1/2" bore, by 1 " rod. The bulk modulus to be used is 200,000 psi. Tube size used were 3/4" OD x 0.065" thickness. Length of tube on blind end is 46.5". The length of tube on rod end is 38.75".

So Start...MAX acceleration...CONSTANT VELOCITY ..MAX Deceleration and stop over the stroke distance of 30".

Regards Woody

This message has been edited. Last edited by: Woodygb,

I believe that you should then calculate the acceleration and velocity using natural freq...I don't know how ... and from this you calc the Force required to accelerate your load...add together Force of Friction and Force of acceleration = Force Total.

Then it's
Force total / area of cap end = pressure
Force total / area of rod end = pressure

The flow rate will be a function of the Velocity and the Cylinder end areas.

Regards Woody

ok ..I've found some relevant info and will try to put it into a spreadsheet.
Would an Excel spreadsheet be of any use to you Maglub?

This message has been edited. Last edited by: Woodygb,

Yes, spreadsheet is ok.

I have answers, just wanted to compare with other calculation process. I am thinking that Peter has different process because the book tried to avoid some highly mathematical calculations and indicated that it can be solve by higher math.

I don't know about harmonics, but the principle was also used aside from bulk modulus and hooke's formula.

I also told Peter in my previous post to use 1/3 of the resultant NF. The most important thing that I would like to know is the calculated natural frequency and from there I could just substitute it to the formula to find all other unknowns.

Here is the calculated natural frequency w=91.4 radians/second and 1/3 of this is 30.46 radians/sec. The acceleration time is .032 sec and the acceleration stabilizing time is .20 seconds(6 times the acceleration time). Hope this helps...maybe you can reverse calculate if the formula has error.


Thanks.

This message has been edited. Last edited by: maglub,

The only fiddly bit... lots of terms... is a math value concerning the volumes. ?

The rest seems to be fairly basic.

Natural Freq = Square root of ...

[ area cap end^2 * bulk modulus] / MASS (lbs-sec2/in.) * ?

PLUS

[ area rod end^2 * bulk modulus] / MASS (lbs-sec2/in.) * ?

MASS = LOAD in lb's / 386.0886 in.sec^2 Which is one Snail or 12 slugs .

I'll try it out tomorrow.

Regards Woody

This message has been edited. Last edited by: Woodygb,

Yes, your right and in the right track...ok.

Here's another clue... Try to find the minimum spring constant and remember since proportional valves meter in and out the flow, sum up all the spring constants involved.



If you get stuck after the calculation of the minimum spring constant, I will give you another clue to proceed...if you needed.

Btw mass stayed in 1000lbs...I have to find out why later...

Gudnite to all...

Good morning,

Maglub SAID

quote:

Btw mass stayed in 1000lbs...I have to find out why later...

NO.
Weight = ( lb's ) and Mass = ( weight in lb's * gravity constant )

Woody.

Damn Woodygb, your too good!!! Your you got it, and my books answers are just a little bit lower but I rather use your answer because the bigger the better. I will note the margin of error so that when I come across any real calculation, I will be safer...

The "d" that I have is 20inches, and yours is around 20.6, that may affect some monitoring instruments location but I am sure that by reducing the tube lengths, the errors can be rectified in actual application without altering the flows. At least if the valve choosen has a greater adjustability range, the better.

Back to work...layers...

Thank you and good morning.

This message has been edited. Last edited by: maglub,