Web tension amatuer. Please verify calcs for dancer?

Hi all,

I will try as best as I can to draw you the mental picture. I am winding flexible plastic film with a gauge of 60 micron and a web width of 400mm. The machine is existing and is 12 years old and has no tension display. It merely shows gauge pressure of the air to the dancer cylinders. I have been tasked with giving the operator an extra field on his HMI showing web tension in Newtons. I intend to keep it simple and just buy a pressure transmitter to feed into a spare analog input on the PLC.

Now for the math before I translate it to the PLC...

I have a dancer in the second order lever configuration. It has two vertical arms hanging down from the fulcrum, 1 on each side of the machine frame. I have two 50mm low friction cylinders mounted 120mm from the fulcrum and I have the dancer roller mounted at 365mm from the fulcrum. The dancer cylinders gauge pressure on the particular job I saw running on the machine showed 6psi. The operator informed me that that is the lowest they run and the highest is 12psi.

So how much web tension is on the web? I have given it a go, so please do tell me if I am wrong.

Find force exurted from cylinders @ 6psi. (Cylinder extending not retracting so I don't have to take into account the piston rod diameter.)
F = P*pi*r^2 [P is in kPa, r is radius in metres.)
F = 81.227N
double that becuase of two dancer cylinders so total force is 162.454N

Find effort force on dancer roller to equal the dancing cylinders force.
Second order lever so a ratio calculation will work.
D[cyl]/D[roller] = F[cyl]/F[effort]
120/365 = 162.454/F[effort]
F[effort] = 53.41N

Find web tension. The wrap angle is 180 degrees at an angle that equally opposes the angle of force of the dancer cylinders

Tweb = F[effort] / (2*(sin[wrapangle]/2))
Tweb = 26.71N

I have ommited loses because I am not an engineer and don't know much about that side of things. Do I need to account for that? So, am I right or close? If i'm not may you please show me why so I can learn some more??? I got all this information from google.

Thanks for helping if you can... Nathan.

If there is a variation - the operator will see something different from normal.
I believe this is the reason that they want a display.
If there is no control adjustment - even if the display was 0~100%
that would be all the operator would need - so long as it was a repetative figure.
If the engineers want it more accurate then ask them

use a weight scale and test it.
the calcs seem alright.

This is stage one of a two stage project. I also purchased an E/P transducer because they want taper tension.
I just wanted to make sure I was calculating tension correctly.
Thanks for your help.