TE Compensation Question

[Hellman]

I've been flying gliders for years, but the following question only
occurred to me recently: In mechanically compensated varios, the TE
pressure is supposed to be as much below static as pitot is above
that. That makes sense since "ram air pressure" increases as v^2, as
does the kinetic energy of the ship. But pitot pressure is
proportional to IAS, whereas the kinetic energy of the ship would seem
to need TAS. If I haven't made a mistake (and please correct me if I
have), TE compensation will only be correct at one altitude. At higher
altitudes it will be under compensated, and at lower altitudes over
compensated. Comments please!

Martin

[Jim White[_3_]]

The TE pressure is not necessarily equal to the difference between static
and pitot. It is unlikely to be nor does have to.

The working principle is that the static pressure changes with altitude,
but as the TE pressure is reduced when you go faster and vice versa, the
two components should cancel each other if you push or pull.

If they do not you have climbed / sunk due to rising / falling air.

Jim



At 01:39 18 June 2009, Hellman wrote:
>I've been flying gliders for years, but the following question only
>occurred to me recently: In mechanically compensated varios, the TE
>pressure is supposed to be as much below static as pitot is above
>that. That makes sense since "ram air pressure" increases as v^2, as
>does the kinetic energy of the ship. But pitot pressure is
>proportional to IAS, whereas the kinetic energy of the ship would seem
>to need TAS. If I haven't made a mistake (and please correct me if I
>have), TE compensation will only be correct at one altitude. At higher
>altitudes it will be under compensated, and at lower altitudes over
>compensated. Comments please!
>
>Martin
>

[Bruce Hoult]

On Jun 18, 1:39*pm, Hellman > wrote:
> I've been flying gliders for years, but the following question only
> occurred to me recently: In mechanically compensated varios, the TE
> pressure is supposed to be as much below static as pitot is above
> that. That makes sense since "ram air pressure" increases as v^2, as
> does the kinetic energy of the ship. But pitot pressure is
> proportional to IAS, whereas the kinetic energy of the ship would seem
> to need TAS. If I haven't made a mistake (and please correct me if I
> have), TE compensation will only be correct at one altitude. At higher
> altitudes it will be under compensated, and at lower altitudes over
> compensated. Comments please!

Of course you are correct that at a given TAS the suction generated in
the TE probe will decrease proportional to the air density at that
altitude.

However don't forget that the rate of air density at altitude does not
decrease linearly.

The air pressure at altitudes we fly at is pretty much 1013 / 2^
(altitude/18000). If you work out the derivative of that you'll find
the rate of change of pressure at a particular altitude is
proportional to the pressure at that height.

That means that the change in potential energy from a 1m change in
height will *also* be misread proportional to the air density at that
height.

So the two effects cancel each other out.

[bildan]

On Jun 17, 7:39*pm, Hellman > wrote:
> I've been flying gliders for years, but the following question only
> occurred to me recently: In mechanically compensated varios, the TE
> pressure is supposed to be as much below static as pitot is above
> that. That makes sense since "ram air pressure" increases as v^2, as
> does the kinetic energy of the ship. But pitot pressure is
> proportional to IAS, whereas the kinetic energy of the ship would seem
> to need TAS. If I haven't made a mistake (and please correct me if I
> have), TE compensation will only be correct at one altitude. At higher
> altitudes it will be under compensated, and at lower altitudes over
> compensated. Comments please!
>
> Martin

I predict this will be a long thread.

First principles:

Total energy vario systems sum the kinetic energy of speed and the
potential energy of height and display the rate at which that sum is
changing with time in knots or meters per second.

This calculation must be done in the same domain and in the same
units. i.e. true airspeed + true rate of climb or indicated airspeed
+ indicated rate of climb in the air-data domain. TE calculations can
also be done entirely in the inertial domain though no such system
exists to my knowledge. Don't mix units or domains.

An example is the Borgelt B50/B500 varios which are altitude
compensated to give true rate of climb and which calculate TE with
true airspeed to produce a fast, smooth and accurate TE vario.

These varios, like most, use a probe which may be regarded as a
special static port which is carefully crafted to respond in a precise
way to airspeed changes while being relatively insensitive to yaw and
pitch angles. The real 'magic' of an air-data TE vario is in the
probe.

You could, if so inclined, connect the TE port on the vario to regular
static ports and have a perfectly workable, if non-compensated, vario.

[Larry[_5_]]

If you accept

0.5*rho*v^2 + rho*g*h + p = const

then the density cancels out when you trade-off a delta_h for a
delta_v^2.

Larry

[Hellman]

On Jun 18, 12:31*pm, bildan > wrote:
> This calculation must be done in the same domain and in the same
> units. *i.e. true airspeed + true rate of climb or indicated airspeed
> + indicated rate of climb in the air-data domain. *TE calculations can
> also be done entirely in the inertial domain though no such system
> exists to my knowledge. Don't mix units or domains.

I think you've have hit the nail on the head. It's annoying that I
missed that since I'd wondered some time ago about indicated vs. true
rate of climb.

The only remaining question in my mind is if IAS and IROC (Indicated
ROC) change in the right manner. At 18k feet where the atmospheric
density is roughly half that at sea level, TAS = SQRT(2) * IAS and
"true kinetic energy" will be twice "indicated KE." For the two
factors to cancel out, it would seem that the TROC (true ROC) would
have to be twice the IROC, and that sounds right. But it would be nice
if you could say a few words about why that's the case. Here's what I
can see right now:

If, as I seem to remember (and makes intuitive sense), pressure vs.
altitude decays exponential, then the derivative of pressure versus
altitude will be half what it was at sea level. If the vario measures
the rate of change of pressure, then we'd get the factor of 2 we need
and all would be well, but ... the capacity bottle has only half as
many air molecules. If that has any impact on the vario (over and
above the factor of 2 mentioned above), then it would seem like
compensation would be altitude dependent. I suspect that it does not,
but any words on why would be appreciated.

Thanks.

Martin

[bildan]

On Jun 20, 12:18*am, Hellman > wrote:
> On Jun 18, 12:31*pm, bildan > wrote:
>
> > This calculation must be done in the same domain and in the same
> > units. *i.e. true airspeed + true rate of climb or indicated airspeed
> > + indicated rate of climb in the air-data domain. *TE calculations can
> > also be done entirely in the inertial domain though no such system
> > exists to my knowledge. Don't mix units or domains.
>
> I think you've have hit the nail on the head. It's annoying that I
> missed that since I'd wondered some time ago about indicated vs. true
> rate of climb.
>
> The only remaining question in my mind is if IAS and IROC (Indicated
> ROC) change in the right manner. At 18k feet where the atmospheric
> density is roughly half that at sea level, TAS = SQRT(2) * IAS and
> "true kinetic energy" will be twice "indicated KE." For the two
> factors to cancel out, it would seem that the TROC (true ROC) would
> have to be twice the IROC, and that sounds right. But it would be nice
> if you could say a few words about why that's the case. Here's what I
> can see right now:
>
> If, as I seem to remember (and makes intuitive sense), pressure vs.
> altitude decays exponential, then the derivative of pressure versus
> altitude will be half what it was at sea level. If the vario measures
> the rate of change of pressure, then we'd get the factor of 2 we need
> and all would be well, but ... the capacity bottle has only half as
> many air molecules. If that has any impact on the vario (over and
> above the factor of 2 mentioned above), then it would seem like
> compensation would be altitude dependent. I suspect that it does not,
> but any words on why would be appreciated.
>
> Thanks.
>
> Martin

I'd say it is worthwhile to have an altitude compensated vario. If
you start with that, then you have to use true airspeed to do the TE
calculation.

More interesting to me is the relatively new Inertial Reference Units
(IRU's) that have been developed for UAV's. These are small, light,
use little power, are accurate and are available at a reasonable
cost. In addition to position data, Euler angles, and 3D
acceleration, they provide very accurate 3D velocity data. The
horizontal and vertical velocity can be used for a nearly perfect TE
vario. It would be immune to gusts and would have instantaneous
response. The vario reading should be very smooth since it's
measuring the actual velocity of the glider without any probes or air
data at all. In addition to perfect vario data, together with true
airspeed, could provide highly accurate real time vector winds.