Recent content by simurq

Thanks! I'll try to replicate what you just recommended. I'll come back...

Yes, the arguments passed to the function are year, month, day, lat, lon, timezone, and dst (1 if on, 0 if off). But again, when the sun/moon is below the horizon (never rises), the function returns a #num, runtime error, which I can trap easily. But to do this "nicely" (showing either "polar...

Sorry if I sound rude to you! It's not my intention anyway... I've developed my app in VBA under MS Excel. And it returns runtime 1004 error failing to properly run the arccos function which, in turn, fails to return a value for latitudes above Arctic circle, that is when the sun almost never...

That's exactly what I would like to know, Sir! The rest I'll do programmatically...

Yes, but it's half of the job - the other half is the season of year which I don't actually know. This would further help with determining polar day and polar night depending on season...

Hello good people, I'm using http://www.ecy.wa.gov/programs/eap/models/twilight.zip by Greg Pelletier to calculate sunrise/sunset times at a desired location. However, the sheet formulae return an error message (#NUM) for latitudes above the Arctic/Antarctic Circle since the sun/moon almost...

Given that an engine's rated power is 160 HP (~119kW), how can I express thrust (T) in Newtons, if aerodynamic drag (D) requires weight expressed in Newtons as well to ensure consistency when calculating power available (Pa) and power required (Pr)?: D = (\frac{1}{2}\rho...

I should have mentioned that by "vehicle" I meant an aircraft. Unlike a car, an aircraft must cope with effects of drag and gravity more than a car mostly subject to ground friction. Therefore, energy saving and particularly the notion of "excess power" is of crucial importance for any aircraft...

Although I'm satisfied with above results, the fundamental question I'd like to ask is: What is the rate of change of POWER with climb? Logically, any vehicle (and aircraft, in particular) climbing up the hill is doomed to lose power directly proportional to the height of climb. And at some...

Correct, thanks for the hint, Sir! But the result is not that different... However: 1010 fpm = 5.13 m/s 78 knots = 40.13 m/s Climb gradient = (5.13 / 40.13) x 100 = 12.79% sin(θ) = sin(12.79) = 0.221301 Therefore, P = 1,338.1 x 9.8 x 40.13 x 0.221301 = 116,457 J/s = ~117 kW Engine's working...

I've come across this discussion while asking similar questions myself re: aircraft climb performance. Pls see attachment at my thread above to visualize the problem. So, judging from your responses and assuming nil winds @ standard ISA conditions (t = 15°C; ρ = 1.225 kg/m3), the power required...