Direct drive minibot - output diameter?

[Tom Bottiglieri]

We did some calculations to find what our desired shaft size was, built a few shafts with diameters a bit over and under that number and tried them all. It turns out in our particular case the base shaft diameter isn't as important as making sure there isn't too much wear on the tread. After it's put it about 10-20 times our tread material becomes noticeably worn and we see drastic changes in the speed.

[pfreivald]

Originally Posted by Ether

May I ask, what is your gear ratio and how much does the bot weigh? I assume you are using 2 motors?

Two motors, gear ratio out of the modified box is 5:1, and the weight I am not 100% sure of, but she's not much more than motors, mounts (with lightening holes drilled), two switches and some polycarb (more lightening holes).

Attached is an Inventor drawing. Ain't she cute?

[Ether]

Originally Posted by pfreivald

Attached is an Inventor drawing. Ain't she cute?

For sure. But how do you deploy it? Is it a press-fit onto the pole, or is it articulated?

[pfreivald]

The polycarb acts as a spring to space out the wheels so they fit around the pole. It's held in place with a bracket on the HOSTBOT which is punched out by a plunger on the MINIBOT when it hits the pole (this plunger also turns on the wheels, and another one turns them back off when it hits the plate).

So the MINIBOT is 'squeezed open' and held in place by a post on the deployment mechanism, then snaps in place and turns itself on using the HOSTBOT-imparted lateral kinetic energy to pop the plunger. It then drives off of the post.

Edit: Attached is a closeup shot (a bit blurry) where you can see the wheels and plunger, as well as an Inventor drawing of our deployment mechanism. (The carriage is deliberately a bit massive. This is not a subtle device!)

[JesseK]

Originally Posted by Paul Copioli

I can't find it now but I made a post a long time ago on this forum (maybe 2004 or 2005) titled "Gear ratio doesn't matter." The point of that post was that gear ratio alone doesn't matter, you have to take wheel diameter into consideration, too. In addition, weight has nothing to do with the optimal gear ratio either. Higher mass means that the wheel diameter for a given gear ratio (in this case a gear ratio of 1) must be smaller to generate enough force to counteract the weight (mass * g) and other acceleration of the mass. To simply show the relationship let's use simple F = ma. F in this case is T/Rw (torque divided by wheel radius) and a is g + your desired acceleration at max power. As you decrease your wheel radius you will increase your force which is needed for higher mass.

The bottom line is that the ability to climb the pole at all really only has to do with the motors and mass (in its simplest form). Motors represent the max available power. You just have to find the right combination to lift the mass.

A more specific relationship than F=ma in terms of rotary acceleration is
Tau = I * Alpha
or
Alpha = Tau / I
Where
Alpha = rotational acceleration
Tau = Torque applied to the rotation
I = Moment of Inertia. Generally that's [mass, in kg] * [radius, in meters]^2

The radius^2 matters in this case. This is really because in almost all cases our sources of force are applied in the forms of torque from a motor. The generic moment of inertia works out because most of the time the mass at the radius far exceeds the mass averaged over the radius (e.g. a wheel weighs much less than the robot it pushes).

Since electric motor torque output is a function of its speed, and speed is a function of acceleration, the overall equation quickly becomes non-linear. So I've only ever put it into Excel to figure out the final numbers. Yet most of the errors I experienced went away when I swapped from trying a direct F=ma calculation to T=Ia (non-linearity induces round off error sometimes, so it's still just an estimate).

[Jeffy]

Originally Posted by Bill_B

.175 OD, right?

Yes. I am satisfying my CD addiction on vacation from an iPad...

[Ether]

Originally Posted by buchanan

The rate of acceleration must be taken into consideration if you want to (and you should) optimize elapsed time rather than simply a top speed there may not be time/distance to reach. Improved acceleration from a smaller shaft trades off against reductions in top speed. This is a somewhat harder problem, but certainly solvable, though I'm personally too lazy to do it.

I set up the differential equation and found an analytical solution. You can see the effect of acceleration, which is interesting and instructive.

As a practical matter other factors argue for a far more conservative value than this optimization would give anyway.

I tend to agree. Compared to variations in motor performance, friction, battery voltage, and whatever else, it's probably not the major factor in determining what the real-world result will be. But still interesting and instructive.

[DonRotolo]

Originally Posted by pfreivald

Nope. No one on the team... And the tech teachers are singularly uncooperative.

Hm, I was thinking of the tech teachers, not someone on the team. This is an odd thing for me to hear, since our tech teachers (and supervisor, and administration, and BoE...) are all very enthusiastic about the robot team. I guess some teachers need to re-learn why they went into teaching...

Originally Posted by pfreivald

This is not a subtle device!)

That's an understatement! Our deployer is similar, riding on some Igus track and using latex tubing as the 'slingshot'. Deploy is about 0.15 second, whacking the minibot onto the pole with audible force. It definitely works.

[AndrewN]

Originally Posted by boomergeek

Has anyone drilled the shafts without using a lathe? I would be curious as to how to get it accurately centered enough and straight in without a lathe.

Using a drill press and a table vice: turn the drill bit upside down and place into the chuck of the drill press, tighten the chuck and make sure it spins without wobble. Turn off the drill press. Lower the spindle so that you can position a vice and clamp the drill bit, clamp the vice into place on the table. Loosen the chuck and raise the spindle. You now have a drill bit clamped into the vice dead center of the spindle. Put the shaft into the drill chuck and tighten. Drill out center of the shaft.

The shaft spins, the drill bit remains stationary.

Take care, plan out all the steps, make sure that the shaft can be placed into the chuck without moving the vice and drill bit.

[boomergeek]

Originally Posted by AndrewN

Using a drill press and a table vice: turn the drill bit upside down and place into the chuck of the drill press, tighten the chuck and make sure it spins without wobble. Turn off the drill press. Lower the spindle so that you can position a vice and clamp the drill bit, clamp the vice into place on the table. Loosen the chuck and raise the spindle. You now have a drill bit clamped into the vice dead center of the spindle. Put the shaft into the drill chuck and tighten. Drill out center of the shaft.

The shaft spins, the drill bit remains stationary.

Take care, plan out all the steps, make sure that the shaft can be placed into the chuck without moving the vice and drill bit.

Thanks for the suggestion. Now we have two.
Your method and teched3's:
"You could try a drill press, drilling a hole in a block of plastic or aluminum with it securely clamped on the drill press table. This hole would be equal to the shaft diameter. Then change drills to a small center drill, and then step up increments to your final hole size. Not perfect, but will get you close."

[Doug G]

Originally Posted by AndrewN

Using a drill press and a table vice: turn the drill bit upside down and place into the chuck of the drill press, tighten the chuck and make sure it spins without wobble. Turn off the drill press. Lower the spindle so that you can position a vice and clamp the drill bit, clamp the vice into place on the table. Loosen the chuck and raise the spindle. You now have a drill bit clamped into the vice dead center of the spindle. Put the shaft into the drill chuck and tighten. Drill out center of the shaft.

The shaft spins, the drill bit remains stationary.

Take care, plan out all the steps, make sure that the shaft can be placed into the chuck without moving the vice and drill bit.

This has to be one of the best fabrication tips I've heard so far. Thank you so much for sharing!

[roystur44]

Here's a tip for those of you without a lathe.

Take the Tetrix wheel hub, remove the set screw and drill the center out to .248" Now the wheel hub will fit right over the pinion gear. Put the locating flange of the wheel hub on the motor side. Add a little Loctite and tighten down the set screw. Now take the axle hub and drill out two of the 6-32 threaded holes(opposite). now you can take two of the allen head screws in the kit and attach the axle hub to the wheel hub. Now you have a nice coupler to attach the axle. Look in the Grainger catalog at shoulder bolts and find one that will fit your specs.

Roy

[scottydoh]

This is what 810 ended up with. Press fit onto the brass spur gear plus a set screw. The actual "drive" part of the shaft measures .287, which happens to be a perfect fit for the surgical tubing we're using.

[philso]

We did something similar to what Roy's team did. Instead of drilling out the motor shaft hub, we hooked the motor up to a bench DC supply and held a file against the pinion gear until it fit in the motor shaft hub.

We put two of the motor shaft hubs back-to-back. We were able to find a pair of holes where the threads on the two hubs for the 6-32 screws matched up without binding on the second hub.

You can then use that to hold pieces of the gear box you have cast aside.

Phil

[jacob9706]

Does anyone know the rpm of the tetroxide motor without the gear box?