# Calculating the driving force in a double-acting cylinder using air

1. Apr 17, 2012

### carle

Hello!

I'm completely lost trying to derive an equation that'll give me the driving force in a DAC, and I really need some help from you guys.

The purpose of knowing this is to optimize the amount of air that is needed (24 litres are available, pressure up to 8 bar) to drive a radio-controlled car (60 kg), which needs 20 N to keep in linear motion.

The connecting rod from the cylinder goes to a scotch yoke, and I guess this is where I get a bit lost because of the force that varies through the angles from 0 to 2π. And I don't quite know when the ideal gas law comes in the picture (it should be used with 1 atm being the ideal pressure at the cylinder bottom, right?).

Data:

* Car wheel radius: 0,1 m
* Force needed to keep the car in motion: 20 N
* Gearing between wheel and engine: 1:6 (that is, one rev for the wheel is 6 for the engine)
* Efficiency is 10-20%
* 24 litres of air available and the pressure is to be choosen between 1-8 bar
* The volume of the cylinder is between 30-40 cm^3 (also to be decided)

My technical english isn't flawless so if something is unclear I'll try to do a better explanation.

2. Apr 17, 2012

### OldEngr63

It is really unusual to build either an engine or a compressor using the scotch-yoke mechanism; the slider-crank is such a much better choice. The scotch-yoke does, however, keep the kinematics simple.

Let x be the piston displacement for crank rotation θ so that
x = R cos θ
where R is the crank radius. This the characteristic of the scotch-yoke.

xmax = x(0) = R
xmin = x(π) = -R
swept vol = area*(xmax - xmin) = 2AR
where A = piston area

The piston does not sweep every bit of the compressed volume. There are always irregularities in the shape of the compression chamber, space around the valves, space in passages ways leading away that contain gas communicating with the compressed volume but beyond the reach of the piston. Taken all together, this is called the clearance volume. It is possible to define a clearance distance, d, such that it looks like additional cylinder length,

clearance volume = A d

With this, the volume as a function of the crank angle looks like

V(θ) = clearance vol + A(xmax - x(θ))

= A (d + R - R cosθ) = A(d + R(1 - cosθ))

If you substitute this volume expression into the ideal gas law (or what ever other pressure-volume relation you propose to use) it will give a means to calculate pressure at any crank angle. As you can see, you will need an estimate for the clearance distance, but there are various ways to obtain that if you think about it.

3. Apr 18, 2012

### carle

Thank you for your reply! I need a little more help though (I'm very new to this); can I use the ideal gas law to calculate the pressure needed to give enough torque for the car to go forward (20N), or how do I do that? In an ideal state, p2 should be 1 atm, but how do I determine v2 and v1? Do I integrate V(θ) = A(d + R(1 - cosθ)) between 0 and π?

I'm also not completely sure how you get V(θ) = clearance vol + A(xmax - x(θ)).

Last edited: Apr 18, 2012
4. Apr 20, 2012

### OldEngr63

I have sent you a pm. Please check it and respond.

5. Apr 21, 2012

### carle

I've replied! Thank you.