Asymmetric Ross Linkage
Posted: Mon Apr 02, 2018 11:27 am
I am designing my first Stirling engine and am going for a Gamma type and I am intending to use a Ross Linkage.
I have been reading Roy Darlingtons book in which on page 30 he outlines the sizing recomendations for a Stirling engine which are:
Power piston diameter = 2x sweep
Displacer swept volume = 1.5 x power piston swept volume
Displacer swept volume = 1/3 total volume
Displacer cylinder length = 3x diameter
On the basis that I want to measure twice and cut once so to speak, I am trying to dome some sums before I start drawing. In order to comply with the dimensions above, everything can be described in terms of the power piston diameter:
So if we take the diameter of the power piston as ādā
Sweep = d/2
Power piston swept volume = pi*d^2/4 *d/2 = pi*d^3/8
Displacer Swept volume = 1.5*pi*d^3/8 or 3*pi*d^3/16
Total volume of displacer cylinder = 9*pi*d^3/16
If we take the diameter of the displacer as āDā then
the volume of the displacer cylinder = pi*D^2/4*3D = 9*pi*d^3/16
Therefore D=0.909*d
The length of the displacer =2.725*d
The crux of all this is that the sweep of the Displacer piston will be 0.908*d whilst the power piston will be 0.5*d. This means that the displacer needs to have 1.82* the movement of the power piston.
All the images that I have seen of Ross Linkages show a symetrical arrangement, but this will not allow for the 1.82 variation in movement. I have put together a spread sheat to calculate the dimensions that I will require for this asymmetrical arrangement and think that it should work but I would be interested in what anybody else thinks.
The spreadsheet can be found below if you are interested.
https://1drv.ms/x/s!AnMn0rJaalvJ-GWjfOTzfe2qIXh7
I have been reading Roy Darlingtons book in which on page 30 he outlines the sizing recomendations for a Stirling engine which are:
Power piston diameter = 2x sweep
Displacer swept volume = 1.5 x power piston swept volume
Displacer swept volume = 1/3 total volume
Displacer cylinder length = 3x diameter
On the basis that I want to measure twice and cut once so to speak, I am trying to dome some sums before I start drawing. In order to comply with the dimensions above, everything can be described in terms of the power piston diameter:
So if we take the diameter of the power piston as ādā
Sweep = d/2
Power piston swept volume = pi*d^2/4 *d/2 = pi*d^3/8
Displacer Swept volume = 1.5*pi*d^3/8 or 3*pi*d^3/16
Total volume of displacer cylinder = 9*pi*d^3/16
If we take the diameter of the displacer as āDā then
the volume of the displacer cylinder = pi*D^2/4*3D = 9*pi*d^3/16
Therefore D=0.909*d
The length of the displacer =2.725*d
The crux of all this is that the sweep of the Displacer piston will be 0.908*d whilst the power piston will be 0.5*d. This means that the displacer needs to have 1.82* the movement of the power piston.
All the images that I have seen of Ross Linkages show a symetrical arrangement, but this will not allow for the 1.82 variation in movement. I have put together a spread sheat to calculate the dimensions that I will require for this asymmetrical arrangement and think that it should work but I would be interested in what anybody else thinks.
The spreadsheet can be found below if you are interested.
https://1drv.ms/x/s!AnMn0rJaalvJ-GWjfOTzfe2qIXh7