Velocities after off center collision

In summary, the wing's linear velocity is reduced when it is applied with a tail plane, due to the additional force needed to keep it in motion.
  • #1
The Sparrow
9
0
Hi, I just have a quick question I hope I can explain well.

Suppose you have a metal rod that you apply a force to for a brief moment at the center of mass. It won't rotate and it will move forward at velocity V. Now you reset the scene and move the the force so it's applied off center off the mass, it will then also provide angular velocity. I'm guessing the rod's linear velocity would now be less, since some of the energy has gone into the rotation. Am I right? and if this is true, how do you go about working out the angular and linear components?
 
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  • #2
I'm guessing the rod's linear velocity would now be less, since some of the energy has gone into the rotation. Am I right?
Edit: Depends on the setup.
and if this is true, how do you go about working out the angular and linear components?
Conservation of linear and angular momentum
 
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  • #3
But how would conservation of momentum equations change? Since I'd do a conservation of momentum for linear, then for angular, but that won't change the outcome of the linear, only the angular.

I'm writing a small program, and I'm adding all forces and torques together before applying the accelerations, and the steps are as follows:

- Work out Force tail plane exerts on plane

- Linear Acceleration = Force/Mass

- Torque = crossproduct of Force and distance from center of mass.

But this method yields the same linear acceleration as if the tail plane were placed on the CG.

What am I doing wrong?
 
  • #4
You are right, the linear velocity depends on the setup:
- If you hit your rod with another object, the linear velocity will be lower, and the colliding object will move a bit (more) in its initial direction compared to a central hit.
- If you apply a fixed force over a fixed time, the linear velocity is the same.

Forces are not a good concept for collisions, however - they are hard to estimate and depend on details of the collision process. Momentum transfer are easier to handle.
 
  • #5
Thanks for the replies guys. I don't have any collisions, its just forces applied to the fuselage. So you say that I can calculate the force that the wing pushes with, and use that same force in both the equations to work out the linear velocity of the plane and the angular velocity? It doesn't seem to be extremely accurate though.

Can I paste the code snippet here so you can see what I'm doing?
 
  • #6
I don't have any collisions
Ok, the topic is a bit misleading then.
It doesn't seem to be extremely accurate though.
Conservation laws are exact.
Can I paste the code snippet here so you can see what I'm doing?
I'm not sure if I can find time for that, but maybe others can help.
 
  • #7
Ok, thank you very much. Sorry about the topic, I thought I could figure it out if I knew the answer to the first question.

Wing.Position is a floating point value that shows how far forward or back the wing is of the CG

SweepMag is the magnitude of the instantaneous velocity of the wing calculated from its angular velocity.

double SweepMag = AngularVelocity*Wing.Position;

vector2df SweepVel = vector2df(0,1)*SweepMag;

// I create the SweepVelocity vector and rotate it by the attitude of the aircraft

SweepVel = SweepVel.rotateBy(Attitude,vector2df(0,0));

vector2df ActualVel = Velocity+SweepVel;

//The actual velocity of the air is the plane's velocity plus the wings angular instantaneous velocity

vector2df Force = Wing.Force(Attitude,ActualVel); //function gives force vector

vector3df ForceArm = vector3df(Wing.Position,0,0);

TotalForce += Force; //linear force += wing force since there are multiple wings

TotalTorque += ForceArm.crossProduct(vector3df(0,Force.getLength(),0));

I'll explain anything in greater detail if needed. And I really appreciate the help. This forum does wonders for the world.

What seems to be happening is that when I only have a tail plane, the aircraft doesn't dive like ad arrow, but gets enough lift from the tail plane to sway left and right in its fall significantly. Thus leading me to think I apply too much force to linear movement when it should almost only be angular.
 
  • #8
I don't see an error in the formulas, but they might be too simple. For the force on a wing, there might be more parameters than just the 2d-motion of its center of gravity (which can be different from the center of drag anyway) and attitude.
 
  • #9
Thanks, I found an error in my code at some other place which corrected itself by causing extreme values. This made the torque dampened out compared to the linear force. It all works now. I realize it's not perfectly accurate, but it's good enough to be fun.
 

Related to Velocities after off center collision

1. What is an off center collision?

An off center collision occurs when two objects collide with each other, but the point of impact is not directly centered between the two objects. This results in a transfer of momentum and energy between the objects, causing them to move in different directions.

2. How do velocities change after an off center collision?

After an off center collision, the velocities of the two objects involved will change. The object that is struck will experience a change in velocity in the direction of the impact, while the striking object will experience a change in velocity in the opposite direction. The magnitude of these changes will depend on the mass and velocity of the objects before the collision.

3. What factors affect the velocities after an off center collision?

The velocities after an off center collision are affected by several factors, including the mass and velocity of the objects, the angle of impact, and the elasticity of the objects. Objects with larger mass or higher velocity will experience greater changes in velocity after the collision, while objects that are more elastic will experience less of a change.

4. How is momentum conserved in an off center collision?

Momentum is always conserved in an off center collision, meaning that the total momentum of the two objects before the collision will equal the total momentum of the two objects after the collision. This can be seen in the changes in velocities of the objects, as well as the direction of their movement after the collision.

5. Can the outcomes of an off center collision be predicted?

The outcomes of an off center collision can be predicted using the laws of physics, specifically the principles of conservation of momentum and conservation of energy. However, the exact outcome will depend on the specific variables involved, making it difficult to predict without knowing all of the parameters of the collision.

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