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--- ideas:super-pressure_balloon_skin_tension_calculations [2018/06/14 21:51] – rocketboy
+++ ideas:super-pressure_balloon_skin_tension_calculations [2018/06/21 12:38] (current) – rocketboy
@@ Line 11: / Line 11: @@
 The local atmospheric pressure, density and temperature can be determined from an atmospheric model.
-If we assume a spherical balloon the skin tension can then be calculated with the standard skin tension equation T = Pd . r / 4
+If we assume a spherical balloon the skin tension can then be calculated with the standard sphere skin tension equation T = Pd . r / 4
 Where Pd is the differential pressure (inside to out) and r is the sphere radius.
@@ Line 34: / Line 34: @@
 So floating at 25,000m the balloon must displace 13.327kg of air – since the local air density is 0.03995 kg/m^3 the balloon will be displacing 333.6 m^3 of air.
-Assume a sphere shape for the balloon balloon - volume V = (4/3) . Pi . R^3
+Assume a sphere shape for the balloon balloon - volume V = (4/3) . Pi() . r^3
 which gives an r of 4.3m – diameter 8.6m
@@ Line 48: / Line 48: @@
 External Pressure Po =  2.48277kPa
-Differential pressure Pd (Pi - Po) = 3.135kPa – 2.48277kPa = 0.6522kPa (about 0.08psi) very similar to observed latex balloon flights.
+Differential pressure Pd (Pi - Po) = 3.135kPa – 2.48277kPa = 0.6522kPa (about 0.08psi - very similar to observed latex balloon flights).
 Surface Tension of a spherical container is given by T = (Pi – Po). r / 4
-Surface tension units are force/length – so the units come out right (since kPa = kN/m^2)
 	T = 0.6522kN/m^2 . 4.3m / 4 = 0.701115 kN/m
-.1N/m = 71.5kgf/m = 157.6lbf/m = 48lb/ft
+.1N/m = 71.5kgf/m = 48lb/ft
-Some of that tension will be borne by the latex balloon, some by the fabric.
+This spreadsheet allows you to calculate the surface tension for a non-eleastic constraining envelope:
+{{:ideas:constrainedfloaterv1.xls|}}