5
$\begingroup$

It seems to me that in this image Neil Armstrong is teaching something related with aerodynamics forces (lift, drag...), but I don't know exactly what is that "Triangle" that He drew (It doesn't seem to be an aircraft).

Neil Armstrong teaching

Neil Armstrong teaching

$\endgroup$
  • 4
    $\begingroup$ Once you get high enough and fast enough, don't worry about lift, just go forward fast and the ground just falls away from you. $\endgroup$ – Robert DiGiovanni May 30 at 22:35
  • $\begingroup$ Yes that is likely it. $\endgroup$ – quiet flyer May 31 at 0:56
  • $\begingroup$ He quite obviously is explaining his findings, that at certain altitude, if your angular speed=the rotational speed of the earth, then your speed equals your drag, and your weight is about half your altitude. This is known as the Armstrong-Stetson -paradox. – $\endgroup$ – Jpe61 Jun 1 at 14:54
  • $\begingroup$ @Jpe61 -- Can't wait to upvote your future answer. $\endgroup$ – quiet flyer Jun 1 at 15:34
  • $\begingroup$ He did fly the X-15, which flew fast enough to take into account the earths curvature. Sort of half flying and half orbiting (which would reduce the lift requirement). Angular speed = rotational speed of earth seems more applicable to geosynchronous orbit. Would have enjoyed his classes. $\endgroup$ – Robert DiGiovanni Jun 1 at 19:00
7
$\begingroup$

enter image description here

1) The triangle shape represents an aircraft-- imagine something like a supersonic version of a Gee Bee racer. The thing at the leading edge of the vertical fin is the canopy.

2) Here is a sketch that offers an educated guess about parts of the diagram that are either cropped, or hidden behind Neil Armstrong, in the original photos. Also, for added clarity, labels have been added to the dashed lines representing the horizon, and the aircraft's longitudinal axis. Note that for the diagram to make sense, the wings must be attached with zero incidence relative to the longitudinal axis of the vehicle, or the vehicle may be a flat-bottomed wingless lifting body. Only the green vectors represent actual forces. "L" is the lift vector, "D" is the drag vector, "W" is the weight vector, and "T" is the thrust vector. Note that "D" is parallel to the flight path, "L" is perpendicular to the flight path, and "W" is perpendicular to the horizon. "T" is drawn in this sketch as nearly, but not quite, parallel to the aircraft's longitudinal axis-- in the original, the direction of the "T" vector is about midway between the direction of the aircraft's longitudinal axis and the direction of the flight path. No attempt has been made in this quick sketch to make the force vectors add to zero. We can't see enough of the original to guess whether the force vectors as drawn there would add up to approximately zero or not.

3) The vector labeled "flight path" in this sketch was labeled "V" (for "velocity") in the original. It represents the aircraft's instantaneous direction (and speed) of travel through the airmass.

4) In the original, the line representing the orientation of the longitudinal axis of the aircraft was simply an extension of the flat bottom of the aircraft; it ended up getting displaced upward a bit in this sketch. This wasn't intentional but does have the effect of causing the line to pass closer to where the aircraft's CG would likely be. The angles of interest can be represented accurately regardless of whether or not the line representing the orientation of the aircraft's longitudinal axis actually passes through the CG or not-- we just need a reference line that is parallel to the aircraft's longitudinal axis.

5) Another detail that ended up different from the original in this quick sketch is the order of the angles illustrated on the left-hand side of the diagram. In the original, running from right to left, the order is gamma, alpha (this letter has been rotated about 45 degrees clockwise so it looks almost like an upside-down gamma), theta, while the order in the attached sketch is theta, gamma, alpha. This has no effect on the meaning of the diagram-- the same greek letters are still associated with the same illustrated angles. A confusing point about the upper photo is that it is cropped in such a way that at first glance the angle labeled "theta" appears to duplicate the angle labeled "gamma", but of course this is not actually the case.

6) It appears that the curved line represents the surface of the earth, the overlined "R" represents the radius of the earth, and "h" represents the aircraft's height above the earth's surface. Neil is probably explaining that at very high speed, orbital effects start to come into play, reducing the need for lift.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Yes, it is rather clear (to me) that the curve is the surface of the Earth (and yes, h is height), and this is the crucial part of the whole thing. Neil is probably explaining that the Earth curvature matters for fast flight (and/or spaceflight). All other things are pretty standard and mundane. $\endgroup$ – Zeus Jun 1 at 0:56
1
$\begingroup$

It appears that he is teaching about the 4 fundamental forces of flight as well as the angle of attack, I also see notes about the thrust angle in the upper image

The triangle is just a crude airplane (I would think). Armstrong had time in a lot of airframes over the years including space craft, this just may be his generic scribble of a "flying machine", it could also just be an X-15 in the first image it appears there is some sketch of the surface of the planet.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.