Timeline for Are jet streams a net benefit in time and fuel savings?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2016 at 10:53 | comment | added | GHB | @JanHudec you convinced me, there is no point in reasoning at "fixed time x" it is only confusing, I edited it so that only the operational way of working is displayed. It should be cleared! | |
| Mar 10, 2016 at 10:52 | comment | added | GHB | Let us continue this discussion in chat. | |
| Mar 10, 2016 at 10:52 | history | edited | GHB | CC BY-SA 3.0 |
deleted 417 characters in body
|
| Mar 10, 2016 at 10:49 | comment | added | GHB | @JanHudec operationally speaking, I agree with you, nobody flies aircraft to get the from A to b in time x. it was only a (maybe wrong) assumption to show the principle that in general the slower you fly the lower the Drag (drag still depends quatraticcally on speed $D=0.5*\rho*v^2*S*Cd$, $q= 0.5*\rho_0*EAS^2 = 0.5*\rho*TAS^2 = $ | |
| Mar 10, 2016 at 10:43 | comment | added | GHB | @JanHudec the equation to get from EAS to TAS is the following: $EAS = TAS * \sqrt{\frac{\rho}{rho_0}}$, $\rho$ being the density of the stream and $\rho_0 = 1.225 kg m^-3$. the Drag Coefficient $Cd$ depends on your lift coefficient $Cl$, which depends on your speed. Plus there are some complex non linear effects, so that at higher speeds you have extra $cd$ (drag) due to wave drag. | |
| Mar 10, 2016 at 10:40 | history | edited | GHB | CC BY-SA 3.0 |
added 936 characters in body
|
| Mar 10, 2016 at 10:30 | comment | added | Jan Hudec | The assumption is not simplification. The assumption is totally wrong. The aircraft will generally fly the same EAS (depending on altitude of the jet stream, the difference from TAS might or might not matter) and accept the resulting flight time. Of course if it fliest the same speed, it has (roughly) the same fuel burn rate, so the time determines fuel consumed. | |
| Mar 10, 2016 at 10:27 | comment | added | Jan Hudec | Ad EAS: In other words, EAS is measure of dynamic pressure, while TAS is measure of actual stream, which is not unrelated, but the relation is quite complicated (of course the wave drag depends on TAS (and temperature), but that is not relevant for transport aircraft. | |
| Mar 10, 2016 at 10:24 | comment | added | GHB | @JanHudec an assumption, is just a simplification to show what factors play a role, in the last paragraph I tried to stress it out, probably not clearly enough, that in practice things are more complex, as you said, buffer etc. | |
| Mar 10, 2016 at 10:20 | comment | added | GHB | @JanHudec Equivalent airspeed (EAS) is the airspeed at sea level in the International Standard Atmosphere at which the dynamic pressure is the same as the dynamic pressure at the true airspeed (TAS) here. The dynamic pressure at TAS and EAS are identical. Drag is proportional to Dynamic Pressure $D= q S Cd$. Q being the dynamic pressure, S the reference area and Cd the drag coefficient. So drag depends on EAS as much as TAS. | |
| Mar 10, 2016 at 10:09 | comment | added | Jan Hudec | And this does not answer the question anyway. At the very least a qualified estimate of how much the consumption is affected is needed. | |
| Mar 10, 2016 at 10:08 | comment | added | Jan Hudec | Your assumption is wrong. Flights are not constrained by time. The schedule has enough buffer that the flight won't be late because of head wind and even if it was, nobody really cares. | |
| Mar 10, 2016 at 10:05 | comment | added | Jan Hudec | Drag depends on EAS, not TAS. There is EAS, $V_Y$, at which the drag is lowest and aircraft usually fly at that speed or only slightly faster. | |
| Mar 10, 2016 at 9:23 | history | answered | GHB | CC BY-SA 3.0 |