Timeline for What produces thrust along the line of flight in a glider?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Mar 24, 2020 at 23:47 | comment | added | Robert DiGiovanni | @enbin zheng agreed | |
| Mar 24, 2020 at 23:43 | comment | added | enbin | @RobertDiGiovanni Instead of looking at L from the ground, L's component is the force pushing the glider. Note that L has no component in the direction of V. When V is pointing down, it is always Wt that pushes the glider. | |
| Mar 24, 2020 at 22:51 | comment | added | Robert DiGiovanni | @enbin zheng it does! But only if vertical drag and vertical lift components = W. When drawn from the aircraft perspective W is "forward", when drawn from the earth perspective, L is "forward" (plane is pointing down). Really 2 ways of saying the same thing. The glide path is the same as the ramp the block rests on, supported by aerodynamic forces. | |
| Mar 24, 2020 at 22:00 | comment | added | enbin | @RobertDiGiovanni I mean it's not L but Wt that pushes the glider forward. | |
| Mar 24, 2020 at 18:51 | comment | added | Robert DiGiovanni | @enbin zheng I enjoy seeing other points of view here, and your writings are helpful. Starting with the block on an inclined plane: only 2 forces, W and D. We put a wheel on the block, D becomes wheel bearing friction F. The component that "pulls" the block will be drawn forward of the block CG, but Wn can also be imagined from the block reference. Note as we tilt the block further forward, Wt gets bigger, until it overcomes D. | |
| Mar 24, 2020 at 9:02 | comment | added | enbin | @RobertDiGiovanni Some people think that there are only $L$, $D$, and $W$ on the glider. This is not entirely correct. Simply put, there are only two forces on the glider: one is aerodynamic $F$, and the other is gravity $W$. $L$ is only the component of $F$ in a direction perpendicular to $V$; $D$ is only the component of $F$ in a direction parallel to $V$. Some people think that $Wn$ does not exist, which is also incorrect. $Wn$ is the component of $W$ in a direction perpendicular to $V$; $Wt$ is the component of $W$ in a direction parallel to $V$. | |
| Mar 22, 2020 at 13:35 | comment | added | quiet flyer | @enbinzheng I will answer you here chat.stackexchange.com/rooms/105834/… | |
| Mar 22, 2020 at 12:06 | comment | added | enbin | @quietflyer According to your logic, D is balanced by the component of gravity. Right? Otherwise, WLD will not have zero net force. | |
| Dec 3, 2018 at 19:42 | history | edited | Federico | CC BY-SA 4.0 |
added 994 characters in body
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| Oct 26, 2018 at 7:58 | comment | added | Robert DiGiovanni | W never changes. L and D and V are determined by elevator trim and will stabilize as the glider does. As far as being "phugoid", my gliders don't do that. I would imagine setting weight too far forward with too strong an elevator trim might induce it. But you are right, this is a stability and control issue. | |
| Oct 26, 2018 at 7:28 | comment | added | quiet flyer | "(best from nose straight down drop)"-- if you want to talk about the complete dynamics involved in dropping a glider with the nose straight down and watching the glider eventually transition into a steady stable glide, it will get much more complicated than we have touched on so far.The pitch "phugoid" will play an important role. Now we're talking about stability and control-- that's a whole different subject from the basic balance of forces in a steady glide. There's no objective criteria by which to say "The vectors and L,D,W triangle are the result, not the cause, of the glide condition." | |
| Oct 25, 2018 at 5:38 | comment | added | Robert DiGiovanni | I hope a good L, D, W diagram can be found. In reality, gravity does run the show. The CG and neutral point, horizontally off set, cause the nose to pitch down once the plane starts falling. The velocity is controlled by the elevator trim. The wing pulls the plane sideways anyway you look at it. (best from nose straight down drop). The vectors and L,D,W triangle are the result, not the cause, of the glide condition. Thanks for a great discussion. L,D,W will be checked out. | |
| Oct 24, 2018 at 4:08 | comment | added | quiet flyer | Don't feel bad, even NASA can't get it right. There is no way the L, D, and W vectors shown in this diagram grc.nasa.gov/www/k-12/airplane/glidvec.html can be arranged into a closed vector triangle. The proportions are all wrong. Why doesn't a few minutes of googling the web turn up a nice diagram where L, D, and W are actually arranged in the form of a closed triangle? Yet I am having trouble finding such a thing. | |
| Oct 24, 2018 at 4:04 | comment | added | quiet flyer | It would be a misconception to think that the LIft vector somehow contains a component that acts opposite to drag. The references to the wing creating forward thrust are inaccurate-- yes the wing's lift has a forward component relative to the horizon, but at the same time the wing's lift vector is completely perpendicular to the drag vector. | |
| Oct 24, 2018 at 3:58 | comment | added | quiet flyer | U is not a force. It is a component of velocity. V is the net velocity and U and w are components. K is the net aerodynamic force and L and D are the only components of that force. Remember, to create a steady motion (constant velocity), there is NO requirement for a net force to be acting along the direction of motion. Rather, there is a requirement that the net force MUST be zero. If what you really asking is "How is drag overcome-- how can the net force along the flight path be zero if there is drag", the answer is Gravity. L, D,and Weight form a closed vector triangle. Zero net force. | |
| Oct 24, 2018 at 2:18 | comment | added | slebetman | One way to get a "feel" of how strong the pitch angle influence motion is to try to push a knife, flat-side-down, through butter (or jelly). If the knife is slightly angled you will find that the butter forces the knife to move forward instead of down | |
| Oct 24, 2018 at 1:38 | history | answered | Robert DiGiovanni | CC BY-SA 4.0 |