what is your response ?
looks like the motor and most of the front of the quad are not over the rear wheels. the driver is leaning forward over the wheels
isn't it the force of gravity generating the moment and the resulting horizontal force the needs to be opposed to balance the moment, counter-acting the motor force toward the right?
gravity is generating a moment
the moment, not the direct gravity force, needs to be balanced/opposed (as well as the vertical force of gravity).
it could be opposed by a horizontal force to the left thru the CG of the body.
consider something heavy with a wheel at one end and that wheel against a wall.
what are the forces at the other end?
one is vertical supporting the weight. but isn't there a horizontal force between the wall and object. the wall is pushing against the object ... and the person holding the object also needs to push horizontally to hold it up
you guys seem to think
the guy on the motorcyle can balance the quad in that position while standing still
that all the weight is balanced over the rear wheels
that it is impossible for the robot to be in equilibrium at any angle except zero.
that if it were at an angle, as in the drawings i thought we were discussing, that there is or can be no force to support the CG of the body except for one directly under the CG
not sure you understand what a moment is, that 2 forces can be at any angle from one another and keep an object pinned at one central point from rotating (think small merry go round)
it must be a miracle that sailplanes can generate a horizontal force due to gravity
this case
in that case, doesn't the CG move away from the wheels? what force causes it to move horizontally
My response is that this is a difficult, transient problem depending on the partiality of the resistance and the slipping. the horizontal force is zero at the top, zero at the bottom, but the circular-ish path of the COG as it rotates around the partially resisting axle towards the ground will have some maximum somewhere as the COG accelerates from x1=Lsin(3) to x2=Lsin(90)-partialSlippage and y1=Lcos(3) y2=Lcos(90). It's a bunch of difficult calc and differential equations that I'm not the least bit interested in. Maybe homework and exercises - How long would it take for an upright rigid body to fall to the ground? - Physics Stack Exchange
...
Now, if the arm at an angle isn't falling due to a horizontal force (no torque applied by a free-spinning motor axle) it is much easier:
The force balancing the torque around the CoG is F_horiz = m*g*tan(theta) by summing moments around B, and the acceleration of the arm in the horizontal direction is a constant a_x=g*tan(theta) or a_x=9.81*sin(3)=0.5m/sec2
If there's something else pushing on the body, do tell.
You had a stronger statement earlier:
You've been saying the wheelie guy isn't CG balanced over the axle. Could he do the same wheelie at constant speed with any angle? 10°? 45°? 85°?
Ignoring wind resistance, at any angle other than 0° (tan(theta)=0) the system will need to accelerate at g*tan(theta) to hold the angle.
Yes. And my biking friend can balance/bikestand his bike with the angle between vertical and the line of its axles at many different angles.
Yes
Yes
Yes. I'm saying it needs acceleration to hold an off-0 angle (ignoring wind resistance)
I think I do. If you take moments around the cental point, any number of forces times their moment arms need to sum to zero to cause no angular acceleration.
No. Sail planes do need wind resistance, and water resistance, and a completely different FBD.
If you are saying I seem to think wrong and believe in miracles, please show your math instead of arguing words about how well your fritzing-like FBD proves you can balance a robot at all angles with zero acceleration.
i give yup
i can be wrong, but not for the reason your arguing
before continuing any discussion, i'd like to make sure you understand what my statements mean instead of arguing what you think they mean
I can be wrong as well.
Am I missing some or torque or force on these FBDs?
or
before answering that, can we agree to try to understand each other's diagrams and statements before trying to shoot them down based on a misunderstanding?
By this:
I think you mean that you can have any constant angle at any relatively constant speed with no appreciable acceleration
This diagram?
and I understand you to mean this:
I do think that diagram would look much the same at 45° and the math would be much easier-- Wouldn't all 6 arrows be the same F=m*g length?
Is that a fair understanding of your diagram?
My point of difference is that the "floor" vector acts on the robot, and the "motor" arrow acts on the floor, and, for the purposes of determining the acceleration of the robot, the equal-and-opposite "motor-acting-on-the-floor" arrow isn't a force acting on the robot. And at 45°, the robot would acclerate to the right at 9.81m/sec2
looking for a yes/no answer?
not just jumping into the same arguments?
yes/no??
Sorry, I was trying to demonstrate by action my attempt at understanding your diagrams and statements, while highlighting what might be my area of misunderstanding. I thought the 'yes' was implied.
apparently you didn't understand my statement quoted above
Dave
ok
don't understand (and would like to)
how about stepping back a bit and consider the two previous post which seem related.
the diagram below shows a body with only 2 CW forces creating a moment that will cause the body to accelerate CW. the CG would move directly downward, no shift to the left or right and the point at the bottom left may move upward before falling and moving to the left
in the 2nd diagram below the horizontal arrow on the left presumably shows the CCW force needed to balance the moment
the diagram doesn't show what generates that force and if supplied wouldn't prevent the body from falling, which is ok
but if is it present, it would accelerate the CG body to the right. but while the CG would move to the right it may not accelerate/displace so much that point (bottom left) may move to the left
because the body is pinned to the wheel, the moment generates a force resisted by the wheel that accelerates the body to the right as it falls
and at some point, because it is pinned to the wheel, there may be a force to the left that retards the acceleration, even making it negative so that the CG maintains the circular arc as it falls instead of moving any further to the right
the above hopefully explains the impact of the moment on the body
does this make sense?
still lots of questions
i'd like to discuss the above
This diagram shows a torque \tau through the axle of magnitude m*p*L*sin(theta) which, summing torques around the axle balances the downward torque from the CoG, resisting the fall of the arm. This is the torque that the wheel applies to the arm through the axle. The equal-and-opposite torque is applied by the arm on the wheel. This diagram assumes the arm angle is held constant by the motor action through the axle, and is intended to calculate the torque required. The wheel has it's own FBD to determine the force of the floor required to oppose this torque.
In the second diagram, it assumes the tilted arm is balanced by a force instead of a torque as above, and calculates that force. If you sum moments around the CoG, the torque from the rightwards force balances the torque form the force upwards from the axle. And since \sum(F_x)<>0 it would accelerate at a=\sum(F)/m. This is a way of calculating what force would keep the arm from falling and keep it in the same exact angular orientation. This horizontal force is applied by the axle of the wheel to the arm will keep the arm from falling, and a free-body diagram of the wheel would have the equal and opposite force applied to its axle by the arm.
to maintain the arm angle, something needs to generate the force to the right, and it's the wheel torquing against the earth.
?? at high speeds, wind could provide an opposing force, but at lower speeds, there's only F=ma inertia to oppose it.
Yes, the rightward force on the bottom of the arm results in acceleration of the arm. And the weightless wheel needs to accelerate in sync with the arm ino order to apply the force.
a moment is like torque. a motor turning an axle produces a moment. (where exactly would the "moment force" be using a motor)?
a single force applied momentarily (impulse) to the edge of a body causes it both to briefly accelerates in the direction of the force, giving it momentum (speed * mass) as well as rotate the body giving it rotational momentum
i don't believe this is correct. gravity adds to a CW moment around the point P lower left and the upward force around the CG also adds a CW moment
i realize now that the upward force only contributes a moment because the diagram doesn't show that P is fixed
since there is no CCW force, the moment is not balanced
any force applied directly to the axle creates zero moment around the axle because the moment arm is zero. the only force generating a moment around the axle is gravity.
i see no force applied by the wheel (is see no wheel)-- in this diagram (why the FBD needs to include the wheel and floor)
it is the horizontal force needed to balance the moment
FBD is applied to the system/assembly. -- if you disagree, no point in further discussion
a professor said an aircraft motor exploded, killed people. they did an FBD analysis and found the assembly wasn't balanced. a part didn't explode, the assembly exploded
bear in mind, force is in units of force, lbF or Newtons. torque is in ft-lbs, N-meters. they are not compatible measures
if you sum around the CG, gravity is no a factor since it's moment arm is zero
because this diagram does not specify that P is fixed, if you sum the force contributing to the moment
Ucw = upward force at pt P equal to weight
Fhor = horizontal force needed to balance the moment
moment = L * ((weight + Ucw) * sin(ang) + Fhor * cos(ang))
if P is fixed vertically and the arm can only rotate around P, then
moment = L * (weight * sin(ang) + Fhor * cos(ang))
if \sum(F_x)<>0 is the moment and it is balanced, I don't see why it would cause either linear or rotary acceleration
ignore wind
the vertical force of gravity is opposed by the balanced moment.
i believe the horizontal force due to the motor is opposed by the moment, Fhor. since the moment is due to gravity, i believe it is the force of gravity turned 90 deg due to the moment that the motor is pushing against and results in no acceleration
i believe that the video of segway that a body (woman) leaning forward, no over the wheels of the segway can be maintained in equilibrium, like the quad wheelie
at least the question in my mind is can it be explained how
seems like there's a lot of misunderstanding, improper use of terminology
Balancing a Two-Wheeled Segway Robot
and it is 2 separate diagrams
I drew two different diagrams: one for a case figuring torque on the motor axle, since you were talking about moments and resisting moments, and one with a force and zero torque at the axle, as it if were a free-swinging-hinge-type inverted pendulum or the point where the arm was balanced against acceleration.
Is point P the axle?
In the first diagram there isn't a CCW force, but there is a CCW torque/moment/couple applied through the shaft of the motor, so the moment is balanced.
If you sum moments around the axle and assume no angular acceleration:
tau + 0*m*g - Lsin(theta) *m*g = 0
If you sum moments/torques around the CoG:
tau - L * sin(theta) = 0
With my second FBD I considered balancing the weight of the CoG with a horizontal force at the axle.
I summed it around B and found Fhoriz like this:
I was summing with moments CCW+, and I think you have a sign error in:
With your terms I think you should have this as:
moment = L*(-Ucw * sin(ang) + Fhor * cos(ang))
.. then
0 = L*(-Ucw * sin(ang) + Fhor * cos(ang))
0 = (-Ucw * sin(ang) + Fhor * cos(ang))
Ucw * sin(ang) = Fhor * cos(ang)
Ucw * sin(ang) /cos(ang) = Fhor
At constant angle, the moments around the CoG/B zero and balanced, but it needs Fhor=m*g*tan(theta) Is there any other horizontal force acting on the arm but Fhor to make \sum(F-x) =Fhor = m*g*tan(theta) Is there another horizontal force acting on the arm?
vs
Heh. Two FBDs--explosion imminent. It also looks like it has an unbalanced force 'f' acting on the wheel at the point of contact, F=ma?. Nice figure though.
One uses free body diagrams to study the interaction between systems/assemblies. If you want to take the wheel and arm together, the internal forces at the axle are internal, and cancel out (newton's 3rd law). If you want to discuss the forces, or torques or motions between the arm and the wheel, you need separate FBDs to determine those reactions. If you just care about the acceleration or not of a tilted arm, you could do just the arm.
On the other hand, If you treat wheel and arm together as a system, ignoring the mass of the wheel, there is the force of gravity pulling the CoG down, and the X and Y forces at the point of contact with the floor. Looking back to your diagram:
I would say that the forces acting on the wheel-arm system are the grey arrows in the vertical, and the rightwards red floor arrow in the horizontal. The leftwards "motor" arrow is the force of the motor acting on the floor. If you sum moments around the axle you get (+=CCW):
-mgLsin(theta) + F_x*r = ???
If we're aiming at 0 acceleration, the RHS=0 and F_x = m*g*Lsin(theta)/r.
If you include the yellow f_motor as a force on the arm-wheel system, then it cancels out f_x in the moment equation and f_x = 0 and the only way you get
-m*g*Lsin(theta) + 0*r = 0 is if theta=0;
If that is what you believe, what angle can be maintained at constant speed? 3°? 10° 45° 85°? Is the angle speed dependent or not? If it is speed dependent, how do you calculate the angle given a certain speed? If it is not speed dependent, what angle of tilt between the point of contact can you maintain at zero speed?
I fully believe you could change tha angle of the Segway handle to arbitrary angles at constant speed, but that it would need definitely need to be balanced by the CoG of the passenger.