Bob Johnson wrote:

As long as we're sitting around the campfire and also to show you I can
go both ways, I 'member a time only a couple of years ago that I
experienced the dreaded aerotow line break at 200 ft and 60 kt over the
outbound fence.

I looked out front and it wasn't too exiting, so I gingerly turned
ninety to the left and the scenery looked some better, but not the
greatest, so I REALlY gingerly gave it another ninety to the left and
was really impressed this time as I found I was perfectly lined up with
the takeoff runway. And I recall I hadn't lost too much of my original
60 kt. Will wonders never cease!!

I thought "OK now God, you've made it possible for me to do this little
magic trick perfectly the first time I tried it in front of all my
friends and hangar bums, give me your hand again and let's try just one
more."

So I pulled spoilers, checked gear down and rolled up to my exact
takeoff spot.

And as I popped the canopy, my good friend who shall remain nameless,
said "S--t!, you've gone and lost us another Tost ring, somebody go back
to the hangar and see if they can find another tow rope". It was his
turn to tow, so I pushed back. Nobody else said a word.

Safety lecture from a dummy follows:

I don't remember to this day why I ninetied to the left. During the
previous year's biannual when Juan Batch pulled the plug on me over his
outbound fence, turning right was the correct choice because in that
direction lay the wind, which blows one back over the airport. This
improves the scenery like you wouldn't believe.

When I tried the trick for real solo, the wind lay to my left. I'd like
to think it was instinct. But I believe it was a coin toss.

Anyway, thank God. And Juan.

It Depends

BJ


This raises the interesting question of the height loss during a 180 degrees
turn in a glider or an airplane with a dead engine. I recently had a dicsussion
about that with a friend who is a power pilot and on this occasion made again a
small computation I had already made on this matter. As I never have seen
these results elsewhere, I think it may useful to show that here. Assume
you fly your turn wit an angle of attack which correspond to the speed V
when flying straight and wings level, and that the vertical sink speed
in the same conditions wuold be Vz, then during this 180 degrees turn
flown with a bank angle phi, the height loss is pi*V*Vz/(g*sin(phi)*cos(phi)),
and the turn is flown at speed V/sqrt(cos(phi)). The optimum (minimal
height loss) is when sin(phi)*cos(phi) is maximum, i.e. phi = 45 degrees,
and the product V*Vz is minimum. A glance on a typical glider polar will
show that this last thing is obtained with V just below min sink speed, but
as it is not easy to find how many below, let's assume the turn is done
at min sink speed, this is not very far from the optimum. For a typical
glider with min sink of .6 m/s at 80 km/h (22.2 m/s) the height loss is
8.5 m, for a typical airplane with min sink of 3 m/s at 120 km/h (33.3 m/s)
the height loss is 64 m. This explains why the 180 degrees turn back to
the runway over the outbound fence succeeds in a glider but not in a power
plane.

In the case mentioned above, the speed (60kt) was far over the optimum,
however the result is as expected not catastrophic. Assuming a bank angle
of 45 degrees, the equivalent speed in straight flight would be multiplied
by 1.18, this gives 26 m/s or 93 km/h. Assuming the sink speed is 1 m/s in these
conditions, we get a height loss of 16.6m. This is for a poor glider (L/D =
26 at 93 km/h).