mark kibort

so, thats the point Carl. those pictures, plus the link to the european GT series , shows the diveplanes, canards, whatever, integrated in the nose of the race cars. focus on some type of splitter that would work and integrate well into a biplane type of dive plane. (canard )

mk

ew928

And those sports cars seem to be all sporting some form of too-dangerously-close-to-the-ground splitters for street use.
Air. Most of the time you don't see it.
And a lot of times it goes where you least expect it to.

soupcan

If you can see air, I gotta start drinking whats in your liquor cabinet.

SeanR

LOL, I like the way you think.

soupcan

Thanks, now I am hungry.

Glenn Evans

My point was, and remains, that an aircraft does not need to sustain a nett downwards force to maintain a steady descent, as was stated by Mark (numerous times) and Jon (whom I know knows better).

What really appears to have confused Mark is "steady descent", which means a descent at a contant rate of vertical speed (ft/sec or m/sec). I certainly did not say that acceleration can occur without a nett force acting on the aircraft. An aircraft does not need to be accelerating to descend, and will not be accelerating for the majority of the time spent in a descent.

My first post, #135 was carelessly worded in an attempt to simplify my point. Lift has a precise meaning in aerodynamics, and I was quite wrong in saying that lift equals weight in a steady descent. As Mark pointed out, drag also has a vertical component in a descent (or in a climb).

My second post, #149, states Downforce is not a term which is defined in aircraft aerodynamics, and I used "nett downforce" to mean the sum of the vertical forces acting on the aircraft ie the vertical components of lift and drag (and thrust, in the case of powered aircraft), and weight. I acknowledge that the use of "downforce" was confusing, and apologise for expecting everyone to understand the point that F=ma and that "a" (acceleration) can be zero in a descent.

Mark certainly is correct when he stated that the discussion was confused by differing definitions of terms. For the sake of correctness, I would like to point out that Jason's (mongoose's) definition of lift is incorrect.

[QUOTE=Mongoose;5923099]Basic aero (non-compressible flow) is fairly simple from a balance of forces view. In straight and level, unaccelerated flight. Lift = Weight, Drag = Thrust. Notice the emphasis. Anytime you change those two conditions, the above equalities break down and you have to use vector addition/subtraction to figure out what is going on. Some things to note about the four forces listed above - Weight always acts towards the center of the earth (i.e. down), lift always acts perpendicular (normal) to the wing (usually up). Drag always acts in the opposite direction from the velocity vector, and thrust always acts in the same direction as the velocity vector (Ok, not always [F-22 vectored thrust] - and we're assuming zero angle of attack for the following discussion)

I had a long discussion with various examples here, but this page explains it better: http://www.auf.asn.au/groundschool/index.html

Lift is the aerodynamic force which acts perpendicular to the velocity vector (and perpendicular to the lateral axis of the aircraft ie it has no lateral component), not perpendicular to the wing. Drag acts in the opposite direction to the velocity vector, so lift and drag, by definition, act perpendicular to each other. Jason's AUF link explains this.

To initiate a descent, the stick or control column is pushed forward to pitch the nose down from its attitude in level flight to reduce lift. While the aircraft's weight exceeds the vertical components of lift and drag, the aircraft will have a vertical acceleration downward and the "g force" (normal to the velocity vector) will be something less than 1 (including possibly zero or negative). The stick is then pulled back - to a position forward of that it was just in in level flight - to stabilise the aircraft at the desired airspeed and angle of descent, when the sum of the vertical components of lift and drag will equal the aircraft's weight. (By the way, "weight" is the name of the force exerted due to the effect of gravity.) The sum of the horizontal components of lift, drag and thrust will be zero. From the aircraft's frame of reference, the weight vector is inclined forwards, and the sum of its component in the x axis (along the direction of the velocity vector) and thrust and drag (by definition, along the x axis or direction of the velocity vector) will be zero.

To sum up, a descent is initiated by reducing lift to produce a nett downwards force (the sum of weight and the vertical components of lift, drag and thrust) which produces a downward acceleration. (To be strictly correct, in accordance with Newton's Third Law, a nett force is balanced by the equal and opposite inertial force due to acceleration; F=ma.) In the majority of descents, this nett downward force is transient, as a steady descent, at constant airspeed and rate of descent, will be established, in which the sum of all the forces, both vertical and horizontal - and acceleration - will be zero.

Glenn Evans

The term "terminal velocity" is used for the velocity at which the weight of a falling object is equal and opposite to its air resistance, so it can accelerate no more. It doesn't really apply to aircraft, as they are limited to a certain dive airspeed or "never exceed" airspeed. While this limitation may be structural (as Mark did, in fact, point out in his post #147), in low speed aircraft it is more likely to be aeroelastic - ie the dynamic interaction of aerodynamic and structural forces - which may result in flutter (the unstable and violent oscillation of control surfaces such as the elevators); divergence (torsional failure of a flexible wing) or control reversal (where a control acts in the opposite direction to that intended).

Higher speed aircraft are limited by Mach number. Subsonic higher speed aircraft are limited by their critical Mach number, at which the flow over a part of the wing becomes supersonic, creating shock waves and causing a rapid increase in drag.

Mark's use of "terminal velocity" reminds me of an anecdote in which the British Royal Aircraft Establishment at Farnborough dived several aircraft to their terminal velocities in 1944. The anecdote is in the book "Spitfire" by Jeffrey Quill, who was the chief test pilot for the Spitfire at Vickers-Supermarine throughout the Spitfire's production life. High speed aerodynamics and compressibility were not well understood in 1944 so, in the course of its research, the RAE decided to measure the terminal velocities of three aircraft of similar maximum (level) speed: the Spitfire (a PR (Photo Recce) Mark XI, a Mustang Mark I (P-51A, I think) and a P-47 Thunderbolt. The Spitfire was fitted with a fully-feathering propeller to avoid overspeeding the engine in the dive; in spite of which two engines were destroyed from overspeeding.

I suspect that the RAE expected that the Mustang, with its laminar flow wing - substantially responsible for its extraordinary range - would prove to be the fastest. In fact, the Spitfire achieved just under Mach 0.9; the Mustang about Mach 0.8 "with the Thunderbolt in the 'also ran' category." The Spitfire's high terminal velocity and Mach number was eventually attributed to its (then) very thin wing: 13 per cent thickness to chord ratio at the root and only six per cent at the tip. These Mach numbers were well in excess of the aircrafts' "never exceed" airspeeds.

The US Army Air Corps tested captured Me262s after the war, and established that the aircraft could exceed the speed of sound in a dive. The USAAC produced a report "Summary Report No F-SU-1111-ND, ME-262 A-1 Pilot's Handbook" dated 15 July 1946, which stated "It is also reported that once the speed of sound is exceeded, this condition [ineffective controls] disappears and normal control is restored." Unfortunately, the German website from where I obtained it no longer exists, but I have a pdf copy. If anyone wants a copy, please PM me.

RyanPerrella

+1

+1

The only thing I can say is that all those cars you pictures clearly use a splitter. The spoiler on the S model 928's doesnt qualify as a splitter though. The use of those planes is to clean the airflow which is increased by the use of such a splitter. So without the splitter, there is no use for a little winglet on the front of the car. UNLESS you can sell them to the "i think they're cool crowd" in which case make em and sell em and call it a day.

Carl Fausett

I SAID (more than once) that I threw the canard onto my 91 just to take a photo of it. I thought it would make more sense than a photo of the canard laying on a table.

This is a mistake I will not make again.

mark kibort

very well said.

mk

[QUOTE=Glenn Evans;5928746]My point was, and remains, that an aircraft does not need to sustain a nett downwards force to maintain a steady descent, as was stated by Mark (numerous times) and Jon (whom I know knows better).

What really appears to have confused Mark is "steady descent", which means a descent at a contant rate of vertical speed (ft/sec or m/sec). I certainly did not say that acceleration can occur without a nett force acting on the aircraft. An aircraft does not need to be accelerating to descend, and will not be accelerating for the majority of the time spent in a descent.

My first post, #135 was carelessly worded in an attempt to simplify my point. Lift has a precise meaning in aerodynamics, and I was quite wrong in saying that lift equals weight in a steady descent. As Mark pointed out, drag also has a vertical component in a descent (or in a climb).

My second post, #149, states Downforce is not a term which is defined in aircraft aerodynamics, and I used "nett downforce" to mean the sum of the vertical forces acting on the aircraft ie the vertical components of lift and drag (and thrust, in the case of powered aircraft), and weight. I acknowledge that the use of "downforce" was confusing, and apologise for expecting everyone to understand the point that F=ma and that "a" (acceleration) can be zero in a descent.

Mark certainly is correct when he stated that the discussion was confused by differing definitions of terms. For the sake of correctness, I would like to point out that Jason's (mongoose's) definition of lift is incorrect.

Bill51sdr

You know Carl, I think this thing has run its course. For the sake of those of us that can't help but look in on the occasional trainwreck, maybe it's time to lock this sucker down. If for no other reason than our sanity.

Randy V

So how many you think you'll sell?

ew928

None.













Till he finds a Carbon Fiber fab guy that don't charge an arm and a leg.

Fabio421

Theres a guy who used to post here ( Tierra Pistris ) who has these things mounted on his 928. But he's a ricer at heart.

RyanPerrella

i think he also has a hatch full of 15" subs and amplifiers too