Tethered DC motor with propeller modelling (vertical movement) problem

terster wrote:

In most environments accessible to earthbound humans, gravity exerts a
constant force downward. (Definition of /down/.) If thrust force is up,
gravity subtracts directly. Wake up!

Jerry
--
Engineering is the art of making what you want from things you can get.
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Does it mean, there is an error in the equation I've derived?
Otherwise, I still don't know, how your answer should help me solve my
problem. As I modelled the system, the Laplace transform of the
equation of motion assuming zero initial conditions would be:

s^2*H(s) = F_thrust(s) / m - g,

where the minus sign before the gravity constant means an opposite
direction. Is the equation wrong? How do I calculate the transfer
function from thrust to altitude?

terster wrote:

What are the units of m? Of g? Do they match? If not, you can't add or
subtract them. Maybe I misconstrue your implied parentheses.

Are Laplace transforms really appropriate for such a simple system?

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

On Fri, 09 Oct 2009 08:17:48 -0700, terster wrote:


You separate your bias thrust from incremental.  You set your bias thrust
equal to the apparent force of gravity; what's left over fits nicely into
a SISO system.

Note that lots of interesting things happen between the drive to the
motor and the thrust -- in the end you'll have to model that, too.

--
www.wescottdesign.com

¯AF¯AF¯AF¯AF¯AF¯AF¯AF¯AF¯AF¯TJerry, that wasn'=
t very helpful

There is an error or omission in your model in that it isn't complete.

Heed what Tim said,  there is no way you can change the thrust
instantly.  The rotor or propeller has inertia and resistance.

This system may have several states.  I bet you end up with one for
vertical velocity and elevation, and another for rotor speed at
least.  Here is a model for the DC motor
http://www.engin.umich.edu/group/ctm/examples/motor/motor.html
You can see the DC motor model adds a current state too.

Is thrust linear with rotor speed?

Peter Nachtwey

There is an error or omission in your model in that it isn't complete.

Heed what Tim said,  there is no way you can change the thrust
instantly.  The rotor or propeller has inertia and resistance.

This system may have several states.  I bet you end up with one for
vertical velocity and elevation, and another for rotor speed at
least.  Here is a model for the DC motor
http://www.engin.umich.edu/group/ctm/examples/motor/motor.html
You can see the DC motor model adds a current state too.

Is thrust linear with rotor speed?

Peter Nachtwey

OP will be lucky if the system linearity is sufficient for a linear
transform approach (although I have seen techniques for dealing with simple
nonlinear systems using laplace). It'll depend on what range of operation
the model has to accommodate, and whether it happens to be working close to
any discontinuities such as stall.

I get the feeling that this could be a fair challenge, unless a lot of
simplifying assumptions are made, which are likely to seriously compromise
the fit.

On Sat, 10 Oct 2009 13:08:22 +0800, Bruce Varley wrote:


¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯TJerry,

I dunno -- if the thing is lifting the engine then it should be at a
fairly linear portion of the thrust vs. motor voltage (or thrust vs.
motor torque) curve.

I've got a control systems trainer that I'm (oh so slowly) developing
that basically consists of a motor on a stick, and sees fairly good
linearizion just by driving the motor with the direction * square root of
the desired thrust.

--
www.wescottdesign.com

Thanks for all the answers. Sorry for my late tnanks, I've been out.