How do propellers exactly work?

#1
Hi,
at the moment I am learning how to build aircrafts (big thanks to @Shade here). What still is a mystery to me, is how to create enough uplift for your machine while maintaining vertical stability but without putting hundrets of propellers onto it (it wouldn't look like a real aircraft anymore).

Questions:
1) Does a propeller still generate lift if you rotate it by 23.06876 degree? It obviously generates more lift if it has an angle...
20170403224533_1.jpg
2) Since it seems like a propeller with no angle still generates lift, is this the same effect like a wing on a real airplane generates lift (through air flowing at the top being faster than air flowing underneath)?
3) If this effect is the reason, why does a wing which is mounted vertically (as a stabilizer) not create a force in left or right direction? It seems strange to me that flat mounted propellers still generate lift, but vertically mounted propellers do not generate a force to the side (induce rolling).
20170403224549_1.jpg

All in all, I would be really curious to know how exactly the aerodynamics work, i.e how to generate enough lift and level out the center of drag effectively (should be approx on the COM at full flying speed, right?).

Thanks a lot!


Regards.


P.S.: The reason for this post is, that I try to build Shade's F-20 timelapse, but my plane needs a really high velocity to be able to take off, where as Shade's version from the Steam Workshop takes off after a few meters. I have the same amount of propellers mounted, all rotated so that they are flat. I can't figure why there is such a big difference.
 
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ITR

l̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ֍̫̜̥̭͖̱̟̟͉͙̜̰ͅl̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ
Staff member
#2
It takes the Inverse Transform Direction of the current velocity, then multiples it with the given "AxisDrag" (not actual drag), which for a propeller is [0,0.015,0] (Meaning it takes only the "up/down" velocity relative to the propeller and divides it on 66.6‾)
Thereafter it takes whichever is lowest between the square of the previously stored "currentVelocitySqr" (not actual Velocity), and the "sqrCap" which is 900, and both stores it in "currentVelocitySqr" for the next frame, and multiplies it with the previously gotten value, and adds it as Relative Force.

The key here, is that when it takes the Inverse Transform Direction, it uses a child of the propeller called "liftNormal", which is has a localRotation of [0,0,0.1999559,0.9798049] (Which is about [0,0,23.06876] in Euler Angles), but when it uses relative force, it uses the propeller itself, meaning it will be rotated around 23 degrees. In addition, it uses the "up/down" of "liftNormal", which is around [-0.3918355,0.9200353,0] (direction it points, not degrees), as opposed to [0,1,0] of the main body of the propeller (if the propeller isn't rotated).
 
#3
Ok, I did get that Inverse Transform Direction takes the propellers current velocity vector and transforms it into the propellers local coordinate system. Which would mean, that we now look through the eyes (coo sys) of the propeller instead of the eyes of the world around it. What confuses me is, that it is not clear to me which object and operation uses which coordinate system in your discription. Especially when you talk about the child "lift normal", does it have its own coo-sys (so different from world and propeller) and if so, is there another Inverse Transform Direction used (which would make sense)?

Also which coordinate system does axis drag use (propeller or liftNormal)?

When I rotate the propeller, do I also rotate its local coo-sys (and those of its children)?


Regards.
 

ITR

l̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ֍̫̜̥̭͖̱̟̟͉͙̜̰ͅl̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ
Staff member
#4
Ok, I did get that Inverse Transform Direction takes the propellers current velocity vector and transforms it into the propellers local coordinate system. Which would mean, that we now look through the eyes (coo sys) of the propeller instead of the eyes of the world around it. What confuses me is, that it is not clear to me which object and operation uses which coordinate system in your discription. Especially when you talk about the child "lift normal", does it have its own coo-sys (so different from world and propeller) and if so, is there another Inverse Transform Direction used (which would make sense)?

Also which coordinate system does axis drag use (propeller or liftNormal)?

When I rotate the propeller, do I also rotate its local coo-sys (and those of its children)?
Regards.
If you rotate, scale or move an object, you change the absolute rotation/scale/position of the children, but not the local ones.

Here's an image for reference: (Note that I may have messed something up, but I don't think I did)

The red arrow is the up direction for liftNormal
The orange arrow is the up direction for the propeller
The blue arrow is the velocity in absolute/world
The light green arrow is what the velocity is relative to liftNormal
The dark green arrow is the "y" coordinate of that (note that I didn't multiply with -0.015 to make it easier to see)
The brown arrow is the force applied to the object in absolute/world

So basically, it treats the vector that is relative to the liftNormal as if it was relative to the propeller.
Some things to note:
* I don't inverse the vector, so the brown vector is the opposite of what it would be in the game
* Though I drag the red vector around in this image, it's the orange one that decides the red one in the game
* The propeller is pointing either towards or away from the screen
 

Shade

Active Member
#5
Heavier plane needs more propeller to counter the gravity. Or you can just add speed so that it generate more uplift.

I don't understand any of these mathematical stuff ITR said, but what I know is:
1. The lift generated is the same no matter how you rotated it, its just the vector angle of that was changed.
2. Propeller at flat angle mostly uses drag to prevent the plane from going down.
3. Flat angle propeller angled that way so that the drag make it alot easier to move forward than going down or sideways, that is why its still viable as vertical stabilizer.
 
#6
@ITR:
This animation is great :).
I try to describe it in my own words. From world perspective the propeller (used as wing) moves in a certain direction with certain velocity. We now look through the eyes of the propeller, which sees the air moving instead of itself (inverse transform). The "trick" at this step is, that we do not use the propellers coo-sys, but the lift-normal-coo-sys, which is perpenticular to the propellers surface. Now taking the y-component of this relative velocity means we now have the velocity component that moves perpenticular to the porpeller geometry, i.e. the component that pushes the wing upwards. This value is now squared to suffice the fact that aerodynamic forces increase with the square of the velocity and also multiplied by a factor which is just to adjust the lift to a reasonable value. Now we have the realitve force in the lift-normal-coo-sys, which is now transformed into the propellers coo-sys.
You said, that the components of this relative force are then put into the propeller coo-sys without doing a transformation. If the calculation works like I described, I don't see the reason for doing this "trick", because it was correct until now and would result in a physically correct force-direction if transformed into the propeller coo-sys.

Btw.: What software did you ise to make this diagramm. Could be handy for doing kinematic sketches.
 
#7
@Shade:
Thank you for clarifying this to me.
As far as I understand it now, the propellers generate lift only through its angle of attack. So they work the same as wing panels with the only difference being that they have an angle of attack when attached "vanilla".
I could figure out why my plane had problems taking off. I thought, that propellers generate lift like real wings, so the plane would take off with enough speed even without angle of attack, but this is not the case. So I moved the back part of my landig gear more to the front near the COM. This allowed me to make the plane tilt backwards using the elevators while on ground already at a very low speed. As soon as the plane had a pitch angle, the propellers had an angle of attack and the plane took off.
 

ITR

l̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ֍̫̜̥̭͖̱̟̟͉͙̜̰ͅl̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ
Staff member
#8
@ITR:
This animation is great :).
I try to describe it in my own words. From world perspective the propeller (used as wing) moves in a certain direction with certain velocity. We now look through the eyes of the propeller, which sees the air moving instead of itself (inverse transform). The "trick" at this step is, that we do not use the propellers coo-sys, but the lift-normal-coo-sys, which is perpenticular to the propellers surface. Now taking the y-component of this relative velocity means we now have the velocity component that moves perpenticular to the porpeller geometry, i.e. the component that pushes the wing upwards. This value is now squared to suffice the fact that aerodynamic forces increase with the square of the velocity and also multiplied by a factor which is just to adjust the lift to a reasonable value. Now we have the realitve force in the lift-normal-coo-sys, which is now transformed into the propellers coo-sys.
You said, that the components of this relative force are then put into the propeller coo-sys without doing a transformation. If the calculation works like I described, I don't see the reason for doing this "trick", because it was correct until now and would result in a physically correct force-direction if transformed into the propeller coo-sys.

Btw.: What software did you ise to make this diagramm. Could be handy for doing kinematic sketches.
If it didn't rotate the force vector slightly, it would only work as a "dampener", and not actually give any forward momentum. That said, it may not need to, but that's what happens :-P

I used GeoGebra
 
#9
Hi,
for my intuition a rotationg propeller would work as a damper if its angle of attack is zero (straight wing). I drew the forces as I imagine them on the sketch below.
20170404200530_1.jpg

Usually the Lift force (dark brown) should also produce un upward momentum. The only reason I could Imagine would be if the there is an additional Drag force (blue) calculated which would cancel out the upward momentum of the Lift force. The rotation of the Lift force would result in a greater part of it generating upward momentum, but the other one should also do this if the transformations are done correctly. If this is the case, then if the lift force coefficent would be increased compared to the drag force coefficient, it should result in an upward momentum.
 
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ITR

l̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ֍̫̜̥̭͖̱̟̟͉͙̜̰ͅl̺̤͈̘̰̺͉̳͉̖̝̱̻̠̦͈ͅ
Staff member
#10
Hi,
for my intuition a rotationg propeller would work as a damper if its angle of attack is zero (straight wing). I drew the forces as I imagine them on the sketch below.
View attachment 1055
Usually the Lift force (dark brown) should also produce un upward momentum. The only reason I could Imagine would be if the there is an additional Drag force (blue) calculated which would cancel out the upward momentum of the Lift force. The rotation of the Lift force would result in a greater part of it generating upward momentum, but the other one should also do this if the transformations are done correctly. If this is the case, then if the lift force coefficent would be increased compared to the drag force coefficient, it should result in an upward momentum.
It wouldn't be un-upward relative to the propeller (Remember that all blocks have to be placed at 0,90,180 or 270 degrees in vanilla)

It's not IRL phyisics, it's how the game is designed
 
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