Chief Delphi

Chief Delphi (http://www.chiefdelphi.com/forums/index.php)
-   Technical Discussion (http://www.chiefdelphi.com/forums/forumdisplay.php?f=22)
-   -   Types of Drives? (http://www.chiefdelphi.com/forums/showthread.php?t=22039)

Jnadke 30-09-2003 13:36

1 Attachment(s)
Quote:

Originally posted by Frank(Aflak)
edit: I looked over your post a second time, and it is true that at the diagonals you get sgrt(2) times the power of each individual set of wheels, but I think you failed to consider that you ALSO get sqrt(2) times the speed, so twice the power. You gotta look at both. And you need a driver who is aware of all of this.
Not exactly true, because velocity is a vector as well, and vectors can be divided into their x and y components. It's simple physics. If you deal with only one direction, then one side contributes sin 0 = 0 to travel and the other contributes sin 90 = 1 times the power, so 0+1 / 2 = 0.5 or 50% of the available power.

Now, if you travel diagonally, the angle of the vector with respect to the axis is 45 degrees. The vector can be split into its forward (y) and sideways (x) components. We have total Power (P), Power in the direction of travel (Py), Power in the sideways directions (Px), Motor power for one set of wheels, MS1, and motor power for another set of wheels, MS2. Py for the left set of diagonal wheels is MS1 sin 45 = P * 0.70. Py for the right side of diagonal wheels is MS2 sin 135 = P * 0.70. Now when you add these, you have to keep in mind that you're only taking into account the power for an individual set of wheels. When you do the multiplication and addition, you can see that you'll end up with LESS power than available.
Then there's the power in the x direction Px = MS1 * cos 45 = MS1 * 0.70 for one set of wheels. For the other set, you have Px = MS2 * cos 135 = MS2 * -0.70. You can see that the power in the x direction is lost because the motors can cancel eachother out.

Keep in mind that power is a function of both SPEED and TORQUE. The same equations can be used for both speed and torque individually. If you want, I can draw some force diagrams later to further prove my point.

Frank(Aflak) 30-09-2003 18:31

say you have two motor sets, as you said MS1 and MS2. Each turns its wheels to give, say, 5 feet/second travel and 10 Newtons of force (numbers off, but to illustrate my point)

in a tank drive setup, the speeds of the motors DO NOT add, so you get a drive that gives 5 feet/second and 20 Newtons of force.

In a four wheel omni setup, the speed increases, and so does the power, so you get 5rt(2) or 7.07feet/second speed, AND 14.14 N of force. Say you adjust the gearing on this omni drive so that instead you get 5.0 f/s diagonal speed. you multiply the gearing by 5.00/7.07 . . . so now you have less speed, but at the diagonals you would have 5 fps.

but you also get more force out of the motors, namely 14.14 N out of each set. make those two legs in a rt triangle and the total is 20N

QED: same sets of motors will give you 5 fps w/ 20N in BOTH tank and Omni configuration.

edit: if it helps, think about it like this: in both setups you have every motor turning as hard as it can. You said yourself (or someone else did) that four wheel omni is not inherently inefficient. Where does that 30% of yours go?

Sachiel7 08-10-2003 15:16

Well, I posted in another thread about this, but I'll put it here, anyway.
Basically, I (and I'm sure many other teams) have been busy this summer designing a Multi-drive system, that functions as Skid steer, Crab, Car, and Angle, all at the simple touch of a button.
The design is set, the parts inventoried, it's been prototyped several times, each a success.
The entire system costs about $800, which includes the Chassis, drive components, and spare parts. It has an estimated weight of about 40lbs w/o electronics. The system also has several backups. For example, if two of our wheel assemblies (somehow) were magically crushed by a couple hundred pounds of force, (I think they'll be as durable as the chassis, and they are well protected) and we don't have enough parts to rebuild them, the system can be converted into a basic skid-steering system by removing parts, and without any reprogramming. This would give us a quick fix at competition if the worst happened.
It is all controlled by 1 joystick, which gives even output to the motors. As well, each wheel assembly is equipped with the proper sensory to detect it's motion, and each pivot drive has the appropriate sensory to detect the wheel's angles. This, accompanied with a Gyro, will give us an edge in auto mode, and we hope to develop a program similar to the StangPS's WildDraw to generate autonomous code.
We're planning ahead this year...
Don't be surprised if our bot is complete before the 3rd week starts!

Frank(Aflak) 08-10-2003 18:31

Let me clarify what I said earlier, with all this algebra and whatnot:
You have two sets of wheels. Each can turn at X rpm's and has a stall torque of Y. using a gear ratio of Z, set giving you X/Z RPM's and Y*Z Nm of torque. With a wheel radius of R, this gives you (2piR)(X/Z) m/s, and (YZ)/R N of push.


Say you arrange the two sets in a tank-style drive. Your top speed DOES NOT INCREASE. Your force is doubled. You now have a top speed of (2piR)(X/Z) m/s and a stall force of (2YZ)/R.

now arrange the sets perpendicular to each other with omni wheels. One set drives the robot in the horizontal direction, the other set drives the robot in the verticle direction. The final speed and power of the robot is the sum of the vectors.

We are looking at the 45deg angle case, where both sets are driving full power. For the sake of simplicity, we look at the right-up case.

the speed vector to the right is this: sqrt(((2piR)(X/Z))^2+((2piR)(X/Z))^2) You understand? Its a right triangle, on leg being the right speed vector, the other the up speed vector. The final speed vector is the hypotenuese. It simplifies into rt(2)(2piR)(X/Z)

Same this with the force vectors. Final result is sqrt(2((YZ)/R)^2)). which also simplifies into rt(2)((YZ)/R)Do you accept this?


Say you adjust the gear ratio (Z) on the omni setup to match the top speeds. the top speed of the omni =sqrt(2((2piR)(X/Z))^2). The top speed of the tank is (2piR)(X/Z). so Z(new), the new gear ratio for the omni platform =


sqrt(2((2piR)(X/Z))^2)
(2piR)(X/Z)

which simplifies into

sqrt(2)
1

so, replace Z in the omni calculations with Zrt(2)

the final speed vector was equal to rt(2)(2piR)(X/Z). Replacing Z with Zrt(2), we obtain (2piR)(X/Z). Fancy that! Exactly the top speed of the tank style drive.

Lets try the stall force now!

rt(2)(YZ)/R.

Replace Z with Zrt(2) and you get 2(YZ)/R

Yah! same as tank-style!

So, in summary.

The tank drive can give you this:
Speed=(2piR)(X/Z)
Stall Force=2(YZ)/R

The omni drive base gives you this:
Speed=(2piR)(X/Z)
Stall Force=2(YZ)/R


where X=free RPM's of motor,
Y= stall torque of motor
Z=gear ratio
R=Radius of wheel.

OneAngryDaisy 08-10-2003 21:11

Re: see if anyone remembers our solutions
 
Quote:

Originally posted by ajlapp
see if anyone can remember our solution way back in 2002.......

no programming needed, no calibration tricky of potentiometers

Oh oh oh! I know!

::Raises hand::

You guys tied three joysticks together at the same configuration of your motors, correct?

SpaceOsc 20-10-2003 12:27

:::blinks:::


All times are GMT -5. The time now is 06:48.

Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2017, Jelsoft Enterprises Ltd.
Copyright © Chief Delphi