Using the load conditions at the beginning of this thread, 12" steel pipe deflects about 6 arc-sec. Too much. So, you either go bigger or just don't bump it in the first place.
This has been gnawing at me. So I redid my numbers. Hopefully this won't be too boring for the denizens of this thread. I thought about sending speedster a PM, but then thought some might find this interesting.
For a cantilevered beam (or column) with a force applied to the end:
d = PL3 / 3EI (sorry, no Greek font, that "d" should be delta)
With a bit of rearranging, since speedster calculated 6" arc (or 0.007", 240" * TAN(6" arc) ):
P = 3dEI / L3
P = 3(0.007 in)(30x106 lb/in2)(279 in4) / (240 in)3 (note, I= 279 is for 12" x 0.375" wall pipe)
P = 12.7 lb (the load at the top of the 20'-0" tall pier pulling sideways).
Since speedster's criterion is 0.5" arc (0.0006 inches) deflection, I reorganized the deflection equation to find I:
I = PL3 / 3dE
I = (12.7 lb)(240 in)3 / 3(0.0006 in)(30x106 lb/in2)
I = 3251 in4
The area moment of inertia for a "hollow" circle is:
I = PI(R4 - r4) / 4
Since, at least in the US, "standard weight" pipe is available, I'll assume the wall thickness will be 0.375":
I = PI(R4 - (R - 0.375 in)4) / 4
Therefore (I wish I hadn't thrown my scratch paper away, I don't remember how I calculated this without expanding the polynomial above):
3251 < PI[ (15 in)4 - (14.625 in)4 ] / 4
3251 < 3829; therefore, 30" x 0.375" wall thickness pipe will be suitable. For reference, the moment of inertia of 24" x 0.375" pipe is 1942 in4, so it isn't stiff enough to meet the 0.5" arc deflection criterion.
Those were the calculations behind my recommendation of 30" x 0.375" pipe for a 20'-0" tall pier with a maximum deflection of 0.5" arc under a 12.7 lb side load. Or, in metric: 762mm x 9.5mm pipe for a 6.1m tall pier with a maximum deflection of 0.5" arc under a 5.8kg side load.