|
|
|
 |
 |
 |
When you are about to enter the course horizontal at top speed, the thrust
and drag are in equilibrium.
The lift and gravity are also in equilibrium.
|
Immediately after the motor off a sharp slowdown will occur because there is no more thrust but only drag still remain. Finally it will stall.
But what effect weight force has when it is in inclined glide slope?
|
|
 |
When glider enter the course with downward angle(a)
Relationship between lift, drag, and weight force are shown in this figure. The force which accelerate glider is the gravity vector W in the direction the glide slope,
It can be expressed as W*sin (a).
@@@@@@@@@ |
L=Lift
D=Drag
W=Weight
a=Glide Angle
d=Horizontal distance or distance flown
Vertical distance or height loss
ksin(a)=Dcos(a)
Lift/Drag=L/D=1/tan(a)=d/h
kcos(a)+Dsin(a)=W
@Drag force equation is@cP^QΟuQrb
|
 |
Letfs calculate L/D of F5B gliders during distance task.
Try a simple calculation. A total flight distance of four laps is 150m X 4 + turning distance.
An average A class flyers distance in turn is about of 5 ~ 7m radius. So r=7. 2Ξr / 2 = 2X3.14X7 / 2 = 22m, 3 turns in one set makes 66m.
But make it 80m because it fly slightly oval, we need to add additional distance.
The average climb height is about 130m for four legs pattern. This makes flight distance of 680m to 130m height loss.
|
@
k^cUWO/PROTDQR
P^TDQRarctan x =PODWX degree @Glide angle is about 11degree
Letfs find out how fast the terminal velocity (Vt) is at 11-degree glide slope.
In most case terminal velocity means falling down speed with W and D force are in equilibriums.
But we could also find out terminal velocity in certain glide angle.
Then we understand when glider is flying faster than (Vt), it is decelerating
to match the (Vt) because of larger D force, and when it is flying slower
than (Vt) it is accelerating to the speed.
The gavity acceleration force from 11 degree slope is W*sin 11degree which
is also equal to Drag force D.
@@@@@@@@@@@vsinPPdegree=P^QΟuQrb
@@@@@@@@@@@uQQvsinPPdegree^Οrb
@@@@@@@@@@@XγiQvsinPPdegree^Οrbj
@@@@@@@@@@@Ο=air densityPDQQT^u@
@@@@@@@@@@@rwing areaODQu
@bcoefficient of dragODOP@When glider is decending at very high speed
there is almost no lift force therefore drag force is also very small.
@@@@@@@@@
@vweightma=massaccelerationPDUXDW/sec2
@V=sqrtQPDUXDW/sec2ODPXP^PDQQT^uODQuODOPSXDSS^
PVW^
@@@@@@ @ uTO/sec@1WO/hr@
@@UWOTO/sec=PRDUsec + 2.5 sec for the climb = 16.1 sec/setBThis matches actual practice time very well.
Let's compare with actual flight data. (Guntamer Rube 2008 Panonia cup)
Full data is shown at How to judge and it's equipment page.
PTth leg@PDXsec@@PTO@VWDX/sec average speed QWS/h
PUth leg@RDPsec@@PVV@TVDP/sec average speed QOU/h
PVth leg@RDTsec@@PVV@TODU/sec average speed 1WQ/h
PWth leg@RDVsec@@PVV@SVDW/sec average speed 1VQ/h
At first glance you could notice small speed drop in each legs. He was flying beautifully with very small deceleration.
You can also guess it's entry speed was about 300km/hr.
It was flying 79m/sec in first leg, more than 150% faster than the terminal velocity, this big difference in speed causes drag to reduce the speed especially at the first turn. After the big loss of the speed at the first turn there is very small decrease in flying speed because actual speed and terminal speeds are very close as long as you can fly cleanly.
The most important point is "Do not dive in the first leg!" it will not accelerate. You only
lose height. You should enter level or even slightly climbing atitudes.
This way your glider turns better at first turn and stii keeps plenty of
height for the second turn, and be able to keep flying downward for 3rd
and 4th legs.
This calculation is to get some idea of what happening during typical F5B
flight and will differ considerably depends on following conditions.
PDentry height@
QDdensity of the air for the day@@@@
RDglide slope angle@
SDflight distance@@@@
|
 |
|
|