Now some considerations about the theory of the equilibrium of the quadcopter.
I’m curious to see if they will fit with the practice.
The force any prop can give is proportional with the square of the angolar speed: Fi=Kw^2
if we call ‘we’ the angular speed that keep the qpi in equilibrium , we get:
Let’s Consider now to increase the w (for a ‘wd’ quantity) for all the motors .What will happen?
So if I want to move up 1m the qpi in 1 second , it means a=2m/s^2. and g=9m/s^2
We get Delta%=sqr(1-2/9)-1 = 0,105.
it is necessary to increase the w by 10% to move up the qpi 1m in un second.
Vertical Equilibrium adding roll or pitch
Let’s assume we want to keep the qpi at a given height but we want to move it horizontally.For exemple we want to roll the qpi.
k cos a (we^2+2wewd+wd^2+we^2-2wewd+2wd^2+2we^2)=mg
cos a (4we^2+2wd^2)/we^2=4
>>>Delta %= wd/we= sqr(2(1-cos a)/cos a)
If I want an inclination of 5′ Delta % = 8% ,so it is necessary to increase by 8 % the w for motor0 and decrease by 8% for motor2
Horizontal equilibrium with roll or pitch
k sen a (4we^2+2wd^2)=ma
mg sen a (4we^2+2wd^2)/we^2=ma
a=g sen a( 1+D^2/2)
with angle a=10′ e D=0.17 the acceleration a=1,58 m/s^2
s=1/2at^2 in 1 second qpi can move about 0,75m
Equilibrio di yaw
To obtain a yaw rotation it is necessary to increase the w for m0 and m2 and reduce it for m1 and m3, in order to keep the equilibrium on the other directions
Let’s say b the distance between the center of the qpi and each motor,and a the rotational acceleration
8bk we wd =Iz a
Delta%=Iz a / (2mgb)
Let’s assume the qpi as 2 crossed bars long 2b (b=200mm) , then
Iz= 2* m/2 * (2b) ^2/12
>>Delta%=b a / 6 g
If we want a rotation on 1 round in 1 sec the angle is
r=1/2 a * t^2 r=2pi , t=1 a =4 pi
Delta = b *4 pi /6 g=2 b pi/ 3 g = 2*200 *3.14 /3 * 9800= 4,2 %