Subject:
|
Re: Mechanical question
|
Newsgroups:
|
lugnet.robotics
|
Date:
|
Wed, 28 Jul 1999 03:55:25 GMT
|
Original-From:
|
Paul Speed <PSPEED@spamcakeAUGUSTSCHELL.COM>
|
Viewed:
|
1162 times
|
| |
|
|
Jon Shemitz wrote:
>
> > Because the wheel base length shrinks in relation to the
> > turning radius. It is this distance that causes the friction. The
> > further this point is from the imaginary circle that the treads
> > describe, the more perpendicular friction there is. As the turning
> > radius increases and the tread length stays the same, the tread
> > more closely approximates this imaginary circle.
> >
> > If necessary I can probably even show it mathematically.
>
> You know, if it wouldn't be a lot of trouble, I'd appreciate it if you
> would. I sort of follow your explanation, but not to the level where I
> could explain this to my 10 yo. I think the math would make it clearer
> to me.
Sure. Based on other comments I've received I'm guessing my
previous explaination was a little obtuse. (No pun intended...
angles, get it.)
Ok, let's see if I can explain this without any pretty
pictures. It is getting to be time for bed so I'm not making any
promises. :)
You have a robot whose treads are some distance apart.
We'll call this D. Half of D is our turning radius R if we
assume we are turning in place. Now, we'll call the length of
the treads L. We'll call the current position within a turn
theta where theta = 0 means that the robot is facing north but
theta actually points east. This means our robot will actually
travel in the direction of theta + 90. This will make it easier
to relate back to the cartesian plane since most of what we are
going to be dealing with is the right-side tread. Are you with me
so far? Ok.
Our treads form a tangent line to a circle whose radius
is R. This is our turning circle. The center (front to back)
of our treads is where the tangent line intersects the turning
circle. Obviously this intersection point will follow the circle
exactly as the robot turns.
As theta changes it is easy to calculate this intersection
point (and therefore the front-to-back center of our treads).
Assume the universe origin is the center of the circle.
xPoint = R * cos( theta )
yPoint = R * sin( theta )
Wheels and treads ideally want to go straight. This is
the direction that they roll and any other direction will drag.
The point where the tangent line intersects the circle is the
only part of the tread that is actually going straight. All
other points are constantly being dragged around the circle to
maintain the tangent line.
To see this in the extreme all you have to do is look
at the front end of a tread. Let's define an imaginary point at
the front of the tread and call it F.
Notice that as the robot turns, F also traces a circle.
However, this circle is wider than the circle defined by radius
R. Why is this? Well, because the center of the rotation will
never change. Both the turning circle defined by radius R and
the new circle that F traces will have the same center. Point
F is farther away from the center than the front-to-back center
of our tread is.
To find the distance use the Pathagorean theorum. Imagine
a right triangle whose base extends from the center of the robot
to the front-to-back center of the tread. The other leg extends
from this point to the tip of the tread (Point F). The hypotenuse
(H) extends from point F to the center of the robot.
So: (please excuse the C notation)
H = sqrt( (R * R) + ( (L/2) * (L/2) ) );
H by its very nature will be longer than R. Draw a
picture if you need to but recognize that since H is a radius
of its own then there is a tangent line too. If we imagine a
tangent line for this larger circle drawn through point F then
we have the angle that F will has to go. Unfortunately, F still
wants to go straight. It will be dragged at the angle defined
by the tangent. The more difference there is in the angle between
this tangent and the direction of the treads, the more dragging
will have to go on.
Because the tread's direction is tangent to the turning
circle it can be expressed as theta + 90. This would be the
direction our robot would travel if we let it go straight.
Likewise, the tangent line that goes through F on the
larger circle will be the angle of H plus 90 degrees. The angle
of H will be:
H Angle = arctan( (L/2) / R )
You would actually need to add theta to be complete, we're
assuming that its still 0 above.
F Tangent Angle = H Angle + 90
(Yes, I can hear a thousand voices screaming into the
night already. :) )
To be really scary, imagine that you have a tread of close
to infinite length. H Angle will approach theta + 90 degrees which
means that F Tangent Angle will approach theta + 180 degrees. Since
our treads angle is actually theta + 90 then the difference between
the angle that the tip of our tread wants to go and the angle it
actually needs to go would be almost 90 degrees. This would be
the same as dragging the tread sideways. Decreasing R to almost
zero would do the same thing.
Now, image you could increase R to almost infinity.
Look back at our H Angle equation. As R increases the angle
of H will decrease. At infinity the angle of H would be the
same as theta and our treads would not drag at all.
CONCLUSION:
So, mathematically we see that decreasing our radius makes
our treads drag more and increasing our radius makes the treads
drag less. Since this is based primarily on the indirect
relationship (L/2) / R then the most significant change happens
quickly as R gets larger. Graph it to see. This means that you
don't have to make robots that are five meters wide... you will
see your most significant improvements long before you get that
wide.
Ok, I hope I've clarified things. I somehow doubt it
though. :) Without a whiteboard to draw on I'm pretty much lost
in this kind of discussion. I could probably draw pictures if
needed.
Good night,
-Paul (pspeed@progeeks.com, http://www.progeeks.com)
P.S.: Just be glad that two dimensions was sufficient for this
discussion. :)
--
Did you check the web site first?: http://www.crynwr.com/lego-robotics
|
|
Message has 1 Reply:
 | | Re: Mechanical question
|
| (...) [lots snipped . By all means, look at the original message! ;-) ] (...) Beautiful stuff! That's what I call an explanation. "progeeks", indeed... :-)) Thanks, Paul! (Incidentally, you're not using a pseudonym? ;-) Best regards, Martin (25 years ago, 28-Jul-99, to lugnet.robotics)
|
Message is in Reply To:
| | Re: Mechanical question
|
| (...) Because the wheel base length shrinks in relation to the turning radius. It is this distance that causes the friction. The further this point is from the imaginary circle that the treads describe, the more perpendicular friction there is. As (...) (25 years ago, 27-Jul-99, to lugnet.robotics)
|
16 Messages in This Thread:
   
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
This Message and its Replies on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
