Squeeze my landing gear

[martins]

Here is a neat little mechanics problem that I thought some of you might appreciate (and that would save me a bit of time, if somebody else solved it):

Given a vessel with mass m and principal moments of ineratia PMI (relative to centre of gravity c=(0,0,0) ), with 3 touchdown points t0, t1, t2, where each touchdown point represents a landing gear whose suspension is modelled as a damped harmonic oscillator,
[math] \ddot{x} = a \dot{x} + b x [/math]with spring constants a and b (which can be different for each gear), by how much would the suspension of each gear be compressed at equilibrium when the vessel is at rest on a horizontal planet surface with surface gravitational acceleration a0?

First prize: 2 days shaved off the upload date of the next beta.

 

[RisingFury]

Latex makes me sad

EDIT:
Ok, I think I fixed the coordinate system so it corresponds with Orbiter's left handed system.

I'll be using Orbiter's left handed coordinate system, with z pointing forward, x pointing right and y pointing up in vessel local coordinates.

The equation for landing gear is given as
[math] \ddot{y} = \alpha \dot{y} + \beta y [/math].

Because you requested a state of equilibrium, we can disregard the second term. That leaves us with

[math] \ddot{y} = \beta y [/math]
Because the vessel is at rest on the ground with constant velocity and constant angular velocity, we can write out the equations for force and moment:

[math] \vec{F_1} + \vec{F_2} + \vec{F_3} = \vec{F} [/math]
where F1, F2 and F3 is the force acting on each landing gear and
[math]\vec{F} = m * \vec{a_0}[/math]

The other equation is for moment:

[math] \vec{M_1} + \vec{M_2} + \vec{M_3} = 0 [/math]
Where M1, M2 and M3 is the moment produced by each landing gear. It equals 0, because the angular momenta are constant.

Where M is defined as
[math] \vec{M_i} = \vec{F_i} \times \vec{r_i} [/math]
I'm making a small error here (don't flog me, please):
[math]\vec{r_i} = \vec{c} - \vec{t_i}[/math]
From that I get the equation:
[math] \vec{F_1} \times \vec{r_1} + \vec{F_2} \times \vec{r_2} + \vec{F_3} \times \vec{r_3} = 0[/math][math] m*\beta_1 * \vec{z_1} \times \vec{r_1} + m*\beta_2 * \vec{z_2} \times \vec{r_2} + m*\beta_3 * \vec{z_3} \times \vec{r_3} = 0[/math]
Where vector Zi describes the landing gear.

Here comes the error. I'm assuming that
[math] \vec{z_i} \times \vec{r_i} = (-x_i * y_i, z_i * y_i, 0) [/math]
From the equations for force and moment, we get three equations:
[math] \beta_1 * y_1 + \beta_2 * y_2 + \beta_3 * y_3 = a0 [/math][math] -\beta_1 * x_1 * y_1 -\beta_2 * x_2 * y_2 -\beta_3 * x_3 * y_3 = 0 [/math][math] \beta_1 * z_1 * y_1 + \beta_2 * z_2 * y_2 + \beta_3 * z_3 * y_3 = 0 [/math]
That solves for:
[math] y_1 = \frac{-a_0 * x_3 * z_2 + a_0 * x_2 *z_3}{(x_2 * z_1 - x_3 * z_1 - x_1 * z_2 + x_3 * z_2 + x_1 * z_3 - x_2 * z_3) \beta_1} [/math]
[math] y_2 = \frac{a_0 * x_3 * z_1 - a_0 * x_1 * z_3}{(x_2 * z_1 - x_3 * z_1 - x_1 * z_2 + x_3 * z_2 + x_1 * z_3 - x_2 *z_3) \beta_2} [/math]
[math] y_3 = \frac{-a_0 * x_2 * z_1 + a_0 * x_1 * z_2}{(x_2 * z_1 - x_3 * z_1 - x_1 * z_2 + x_3 * z_2 + x_1 * z_3 - x_2 * z_3) \beta_3} [/math]
I hope I solved these correctly. The different betas make the formulas clumsy so someone should check my math and methods...

If you want to eliminate the error caused by the definition
[math]\vec{r_i} = \vec{c} - \vec{t_i}[/math]than all you need to do is project the vector ri onto the plane created by points t1, t2 and t3. That will eliminate the error...

 

[martins]

Thanks for that, your Ansatz looks good, this is definitely on the right path.

However ...

Your solution seems to be independent of mass m. (m drops out somewhere along the derivation when you substitute the individual Fi into the equation for the moment.)

This doesn't sound right. Surely the solution must depend on m?

 

[RisingFury]

Hmmmm... hold on, I'll go over my calculations again.

Seems intuitive that the position depends on mass, yes. I must have dropped it out somewhere accidentally.

EDIT:

Here is where I lost the mass:
[math] m * \vec{\ddot{y}} + m * \vec{\ddot{y}} + m * \vec{\ddot{y}} = m * \vec{a_0} [/math][math] m*\beta_1 * \vec{z_1} \times \vec{r_1} + m*\beta_2 * \vec{z_2} \times \vec{r_2} + m*\beta_3 * \vec{z_3} \times \vec{r_3} = 0[/math]
As you can see, in both equations I used m and it drops out. That must be a misinterpretation on my end. I assumed
[math] \vec{M_i} = m * \beta_i * y_i [/math]
However, that's a mistake. It's supposed to be
[math] \vec{M_i} = m_i * \beta_i * y_i [/math]
where
[math] \sum{m_i} = m [/math]
Give me some time and I'll reformulate my solution.

EDIT: This time I used the orbiter coordinate system from the start of the calculation and accounted for the mistake I made before. This time I get these four equations:

Mass:
[math] m_1 + m_2 + m_3 = m [/math]
Force:
[math] m_1 * \beta_1 * y_1 + m_2 * \beta_2 * y_2 + m_3 * \beta_3 * y_3 = m * a_0 [/math]
Moment:
[math] m_1 * \beta_1 * y_1 * z_1 + m_2 * \beta_2 * y_2 * z_2 + m_3 * \beta_3 * y_3 * z_3 = 0 [/math][math] m_1 * \beta_1 * x_1 * y_1 + m_2 * \beta_2 * x_2 * y_2 + m_3 * \beta_3 * x_3 * y_3 = 0 [/math]
The problem now, is that there are 4 equations and 6 variables. A new link is needed.

The link is quite easy to make in one dimension, but seems a bit more difficult in two dimensions.

You can probably simplify the problem if you formulate your problems as
[math] F = \alpha * \dot{y} + \beta * y [/math]because it drops the ratios of masses right out of the equation, but I'll continue to work on this problem anyway.

EDIT:
Right now I'm thinking that maybe it's possible to solve the problem by manually disassembling the problem into two 1D problems.
You get one equation for mass, one equation for force and two equations for torque, one for z dimensions and one for x dimension. I'm hoping that's correct.

EDIT:
Nope, I need another way of figuring out what the ratios of m1, m2 and m3 are. But I've been at it for hours and am getting tired, so I'll continue this tomorrow.

---------- Post added at 22:11 ---------- Previous post was at 18:09 ----------

I've given it more thought. The only way this problem will be solvable is to be able to somehow get the mass ratios m1 : m2 : m3.

The equations I derived seem to work, but that link is missing. You have two options.

You can either change the problem to F = a * y' + b * y. That way you eliminate 3 variables and the problem becomes solvable.

The other solution is to somehow get the mass ratios. Either assume some sort of distribution offhand, or retrieve the data from the mesh with that mesh tool...

I haven't studied this, but maybe the PMI could be a hint towards mass distribution. Maybe ratios could be derived from there...

 

[Hielor]

I'm probably missing something here, but given that you know the center of mass of the vessel, couldn't you just take the mass over each gear as being proportionate to that gear's distance to the center of mass (possibly in just the horizontal plane...)?

e.g., if you had three touchdown points equidistant from the center of mass, then each touchdown point would have an equal share of the load. If you have a mass distribution which would increase the load on one or two of the touchdown points, wouldn't it also change the center of mass?

Odd cases might happen if the center of mass is "outside" the triangle defined by the gear, but in that case, unless the vessel is strapped down somehow it should tip over anyway...

 

[RisingFury]

I think you're probably right about that. I completely forgot about the center of mass.

That gives you the ratios of masses each gear has to carry which solves the problem.

 

[martins]

Not quite. If all touchdown points are equidistant from the centre of mass, but two of them are next to each other, while the third is on the opposite side (notwithstanding stability problems of such an arrangement), then the two points in close proximity would would each only have half the load of the third. So the geometry does matter.

 

[martins]

You can either change the problem to F = a * y' + b * y. That way you eliminate 3 variables and the problem becomes solvable.

So you are saying that you have a solution if all three points are using the same spring constants? In that case, the solution of the general case of having different values of b for the three points should actually be trivial, because they (nearly) separate. For each individual landing gear, there is a linear relationship between b and displacement y, given from the equation of motion, y'' = by. So if you find displacement y for some b, you can directly get the solution y2 for any other b2 by y2 = y b/b2, independent of the other two points. (Assuming that you wouldn't introduce a significant tilt that shifts the relative mass balance.) Or am I missing something here?

 

[RisingFury]

So you are saying that you have a solution if all three points are using the same spring constants?

No, if you assume F = a * y' + b * y, you simplify the math behind the problem to the point where you no longer care about mass distribution. It's a cheap trick.

[math] \vec{F} = \alpha * \dot{y} + \beta * y [/math]
But y' = 0, so

[math] \vec{F} = \beta * y [/math]
The equation for force:
[math] \vec{F_1} + \vec{F_2} + \vec{F_3} = m * \vec{a_0} [/math][math] \beta_1 * y_1 + \beta_2 * y_2 + \beta_3 * y_3 = m * a_0 [/math]
The equations for moment:
[math] \vec{M_1} + \vec{M_2} + \vec{M_3} = 0 [/math][math] \beta_1 * y_1 * z_1 + \beta_2 * y_2 * z_2 + \beta_3 * y_3 * z_3 = 0 [/math][math] \beta_1 * x_1 * y_1 + \beta_2 * x_2 * y_2 + \beta_3 * x_3 * y_3 = 0 [/math]
Solved for y1, y2, y3:
[math] y_1 = \frac{-a_0 * m * x_3 * z_2 + a_0 * m * x_2 * z3}{\beta_1 (x_2 * z_1 - x_3 * z_1 - x_1 * z_2 + x_3 * z_2 + x_1 * z_3 - x_2 * z_3)} [/math][math] y_2 = \frac{-a_0 * m * x_3 * z_1 + a_0 * m * x_1 * z_3}{\beta_2 (x_2 * z_1 - x_3 * z_1 - x_1 * z_2 + x_3 * z_2 + x_1 * z_3 - x_2 * z_3)} [/math][math] y_3 = \frac{-a_0 * m * x_2 * z_1 + a_0 * m * x_1 * z_2}{\beta_3 (x_2 * z_1 - x_3 * z_1 - x_1 * z_2 + x_3 * z_2 + x_1 * z_3 - x_2 * z_3)} [/math]

The reason you can solve it without mass distribution for this definition of the problem is because you already know the force acting on the body.

 

[Miner1]

There's another way to solve this problem that may or may not be more elegant but will certainly generalize the solution if more than three supports exist.

Consider a 2D equavalent problem (extension to 3D is straightforward) of a beam loaded with arbitrary masses and supported by multiple compressible supports modeled as springs. See attachment for a somewhat crude sketch. The distance between each support and the CG, [math]r_i[/math], is known as is the spring constant, [math]k_i[/math], (elastic modulus) of each support. The mass and CG of the beam is also known.

In equilibrium, the sum of forces and sum of moments are zero, this yields

[math]\sum F_z = \sum_{i=1}^n k_i \Delta z_i - mg = 0[/math][math]\sum M = \sum_{i=1}^n r_i k_i \Delta z_i = 0[/math]
where [math]\Delta z_i[/math] is the (downward) displacement of the [math]i^{th}[/math] support. We thus have 2 equations for [math]n[/math] unknowns. If we limited ourselves to only two supports, there would be no problem. But, if we treat the beam as a rigid body, then it will experience a single displacement, [math]\Delta z[/math], and a single rotation, [math]\theta[/math], about its CG. The individual displacements at each support can then be written as

[math]\Delta z_i = \Delta z + \theta r_i[/math]
Using this expression in the force and moment equations above, we get

[math]\Delta z \sum_{i=1}^n k_i + \theta \sum_{i=1}^n k_i r_i = mg[/math][math]\Delta z \sum_{i=1}^n r_i k_i + \theta \sum_{i=1}^n k_i r_i^2 = 0[/math]
Which constitutes two equations and two unknowns. These type of problems are known in the engineering community as statically indeterminate since you can't solve them by statics alone. The solution is found by considering the body itself. We used a rigid body here, but you can also use finite element models or some other elastic model for the body. Hope this helps or is at least just interesting.

 

[RAF92_Moser]

I believe I have a simplified solution for this system, assuming this aircraft is using a tricycle landing gear. This was a nice little mechanics problem. RisingFury did a good job on this problem and Miner1. This solution should be easy to follow.

Simplifying this solution relies on assuming that the back landing gear share equal loads, and that they are symmetric to the z-axis. Also, the forward landing gear will be located on the z-axis. This configuration should be correct for tricycle gear.

Like before we sum up all the forces and moments to zero, since the system is in equilibrium.

[math]\sum\overrightarrow{F_{i}}=\overrightarrow{F_{0}}+\overrightarrow{F_{1}}+\overrightarrow{F_{2}}-\overrightarrow{F_{w}}=0[/math]
[math]\sum\overrightarrow{M_{i}}=\overrightarrow{M_{0}}+\overrightarrow{M_{1}}+\overrightarrow{M_{2}}=0[/math]
The combined force of each landing gear should support the weight of the aircraft.

[math]\overrightarrow{F_{0}}+\overrightarrow{F_{1}}+\overrightarrow{F_{2}}=m\overrightarrow{a_{0}}[/math]
Using RisingFury's great assumption, the force of each landing gear at equilibrium will be:

[math]\overrightarrow{F_{i}}=\alpha_{i}y_{i}\hat{y}[/math]
where [math]\alpha_{i}[/math] is the spring constant (Hooke's Law).

The moment is found by:

[math]\overrightarrow{M_{i}}=\overrightarrow{F_{i}}\times\overrightarrow{r_{i}}[/math]
where r is the vector from the CG to the landing gear.

[math]\overrightarrow{r_{i}}=\overrightarrow{t_{i}}-\overrightarrow{c}=\overrightarrow{t_{i}}=x_{i}\hat{x}+y_{i}\hat{y}+z_{i}\hat{z}[/math]
The moment will be solved by the determinant of this matrix:

[math]\overrightarrow{M_{i}}=\overrightarrow{F_{i}}\times\overrightarrow{r_{i}}=det\left[\begin{array}{ccc} \hat{z} & \hat{x} & \hat{y}\\ 0 & 0 & F_{i}\\ z_{i} & x_{i} & y_{i}\end{array}\right][/math]
Forgive me for showing a lot of work. If I make a mistake, I want people to see it.

[math]\overrightarrow{M_{i}}=\hat{z}(-F_{i}x_{i})-\hat{x}(-F_{i}z_{i})=F_{i}z_{i}\hat{x}-F_{i}x_{i}\hat{z}=F_{i}(z_{i}\hat{x}-x_{i}\hat{z})[/math]
The moment can finally be evaluated as:

[math]\sum\overrightarrow{M_{i}}=F_{0}(z_{0}\hat{x}-x_{0}\hat{z})+F_{1}(z_{1}\hat{x}-x_{1}\hat{z})+F_{2}(z_{2}\hat{x}-x_{2}\hat{z})=0[/math]
Once again, the forces of the landing gear balance the weight of the aircraft.

[math]\overrightarrow{F_{0}}+\overrightarrow{F_{1}}+\overrightarrow{F_{2}}=m\overrightarrow{a_{0}}[/math]
The back landing gear share equal loads due to symmetry.

[math]\overrightarrow{F_{0}}=\overrightarrow{F_{1}}[/math]
The forces can now be written as (direction cancels out):

[math]2F_{1}+F_{2}=ma_{0}[/math]
The moment can now be written as:

[math]F_{1}(z_{0}-x_{0})+F_{1}(z_{1}-x_{1})+F_{2}(z_{2}-x_{2})=0[/math]
[math]F_{1}(z_{0}+z_{1}-x_{0}-x_{1})+F_{2}(z_{2}-x_{2})=0[/math]
By symmetry, the x-components of the back landing gear cancel out.

[math]-x_{0}-x_{1}=0[/math]
And the x-component of the forward landing gear is zero:

[math]x_{2}=0[/math]
The moment becomes:

[math]F_{1}(z_{0}+z_{1})+F_{2}(z_{2})=0[/math]
And F1 can be written as:

[math]F_{1}=-\frac{F_{2}z_{2}}{z_{0}+z_{1}}[/math]
Substituting the previous expression into:

[math]2F_{1}+F_{2}=ma_{0}[/math]
results in:

[math]-\frac{2F_{2}z_{2}}{z_{0+}z_{1}}+F_{2}=ma_{0}[/math]
Substituting Hooke's Law for F2:

[math]\alpha_{2}y_{2}\left(1-\frac{2z_{2}}{z_{0}+z_{1}}\right)=ma_{0}[/math]
And now y2 can be determined in terms with givens:

[math]y_{2}=\frac{ma_{0}}{\alpha_{2}}\left(1-\frac{2z_{2}}{z_{0+}z_{1}}\right)^{-1}[/math]
F2 can be written as:

[math]F_{2}=-F_{1}\frac{z_{0}+z_{1}}{z_{2}}[/math]
And as previously:

[math]2F_{1}-F_{1}\frac{z_{0}+z_{1}}{z_{2}}=ma_{0}[/math]
[math]F_{1}\left(2-\frac{z_{0}+z_{1}}{z_{2}}\right)=ma_{0}[/math]
[math]\alpha_{1}y_{1}\left(2-\frac{z_{0}+z_{1}}{z_{2}}\right)=ma_{0}[/math]
Y1 and Y0 (since F1 = F0) can be now found:

[math]y_{1}=y_{0}=\frac{ma_{0}}{\alpha_{1}}\left(2-\frac{z_{0}+z_{1}}{z_{2}}\right)^{-1}[/math]
This is not a general solution, just a solution for a symmetric tricycle landing gear.