quadcopter

Now some considerations about the theory of the equilibrium of the quadcopter.

I’m curious to see if they will fit with the practice.

Vertical equilibrium

F0+F1+F2+F3=P

The force any prop can give is proportional with the square of the angolar speed: Fi=Kw^2

4kw^2=mg

if we call ‘we’ the angular speed that keep the qpi in equilibrium  , we get:

>>K=mg/4we^2

Let’s Consider now to increase the w (for a ‘wd’ quantity) for all the motors .What will happen?

F0+F1+F2+F3=mg+ma =4kw^2=4mg(we+wd)^2/4we^2

mg+ma=mg (we^2+wewd+wd^2/we^2

gwe^2+wewdg+wd^2g=we^2g+we^2a

gwd^2+gwewd-we^2a=0

>>wd=[-2gwe+_ sqr(4g^2we^2+4gawe^2)]/2g

that meas:

>>>Delta %=wd/we=sqr(1+a/g)-1

s=1/2at^2

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.

F0=k(we+wd)^2

F2=k(we-wd)^2

F1=F3=k(we)^2

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

M0+M1+M2+M4=0

Mi=l*Fi=bk we^2

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

2bk(we^2+2wewd+wd^2-we^2+2wewd-wd^2)=Iz a

8bk we wd  =Iz a

with k=mg/4we^2

2bmgwewd/we^2=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 %

 

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