# Why does pitch in a helicopter take effect 90 degrees later?

In a helicopter if you want to give it a forward pitch, you change the angle of the blades when it is in this position ---- So the two blades experience unequal lift and because o gyroscopic precession the helicopter pitches forward (instead of pitching sideways, which is intuitively expected)

Up till this it's clear to me.

But what I dont understand is, why does this motion have to take effect 90 degrees later when the orientation of the wings is like this --- I got these images from the helicopter physics videos of smarter every day. Please watch this video to understand my question better.

It says that if the helicopter wants to pitch forward, it changes the angle of the blades 90 degrees before. And gyroscopic precession takes place. But it also says that the effect takes place 90 degrees later. That I do not understand. The effect should take place immediately.

• The cyclic control description here was pretty clear to me en.wikipedia.org/wiki/Helicopter_flight_controls Apr 13, 2014 at 9:06
• Could it simply be that you want the helicopter blade to "bite" for the full 180 degrees centered around the perpendicular position? (by "perpendicular position" I mean where the blade is perpendicular to the aircraft)
– user12029
Apr 13, 2014 at 9:15

If you turn an electric fan on, on a table, and throw a ping pong ball past it, you see the air push the ping pong ball to a different direction than you threw.

You throw it one direction, it gets pushed another direction, and it ends up going to a combination of the two directions. Physicists call this vector addition because both the speeds AND directions seem to get added together.

Imagine helicopter blades spinning so fast that they become a blur. It looks like a disc and is referred to by helicopter pilots as the rotor disc.

Hopefully this picture helps you see that just like the air pushes the ping pong ball to a different direction, the force on the rotor disc pushes the motion of the rotating disc to a different direction The Vs stand for "Velocity", which is a name for speed when you include a direction. F is for the "Force" of our push.

(1.) Is a top down view of a counterclockwise rotor disc, like the Robinson R44. (1.) is to help orient us to this 2-dimentional picture of our 3-dimentional world.

(2.) Is a side angle view of the same rotor disc, with a force pushing up on the side closest to us. This force causes the disc to change to (3.)

(3.) Shows that although the disc didn't turn the direction most of us expected, the part of disc we pushed changed direction exactly like the ping pong ball would have. It was going one direction, it got pushed another direction, and it ends up going to a combination of the two. Notice V2 and V4 did not change direction. V2 is still going away from us and V4 is still coming toward us. V1 was going right, it was pushed up, and it ends up going up and to the right. Because the center of the rotor disc is attached to a drive shaft, pushing up on our side causes the other side to be pushed down, and the direction of V3 changes downward.

ABOUT THE OTHER ANSWERS: user10851's answer is correct, but anyone who isn't familiar with physics (and doesn't have awesome spatial awareness) won't understand it. Also, his answer basically says that it happens because the cross product tells it to, which isn't very helpful. NeuroFuzzy and HelloGoodbye are incorrect.

Let's assign some axes. Say $z$ points up through the helicopter, $y$ forward, and $x$ to the right. And let's agree that the blades are spinning counterclockwise when seen from above. That is, the angular frequency $\vec{\omega}$ and angular momentum $\vec{L}$ of the blades are initially both in the $+z$-direction, and the helicopter is hovering motionless.
Case 1: The swashplate is adjusted so that the angle of attack is greater on the left and lesser on the right. On the left ($-x$) there is an upward force ($+z$), so there is a torque on the blade system in the direction of $(-\hat{x}) \times (+\hat{z}) = +\hat{y}$. On the right side we again get the same torque, as $(+\hat{x}) \times (-\hat{z}) = +\hat{y}$. As a result $\vec{L}$ is shifted from $L_0 \hat{z}$ to something like $L_0 \hat{z} + L_1 \hat{y}$, where $L_0, L_1 > 0$. This corresponds to being "tilted forward" such that there is now thrust driving the craft forward.
Case 2: Now let's again start at hovering and increase the angle of attack ($+z$ force) in the front ($+y$ displacement) and decrease it ($-z$ force) in the back ($-y$ displacement). The same cross-product induced $90^\circ$ phase shift occurs, as $(+\hat{y}) \times (+\hat{z}) = (-\hat{y}) \times (-\hat{z}) = +\hat{x}$. The $+x$ torque shifts the angular momentum to be about an axis tilted up and to the right from pure $+z$, and the helicopter moves right.
As for where/when the effect takes place, there is something of a subtlety. Certainly any impulse on the blade will be transmitted instantly in time throughout the body (or at least at the sound speed inside the material). But the effect of that impulse is mostly to alter the trajectory of the blade. The faster the blade is spinning the further the vertical apex of the new trajectory is carried toward $90^\circ$ downstream. The thing about helicopters that trips up intuition is the fact that blade angles alter forces directly but not displacements.