Chief Delphi - wiring

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- Motors (http://www.chiefdelphi.com/forums/forumdisplay.php?f=52)

- - wiring (http://www.chiefdelphi.com/forums/showthread.php?t=93978)

Quote:

Originally Posted by clayman (Post 1045172)

Conclusion: it won't work :rolleyes:

we could always play next year's game on a sloped field and use gravity...

Quote:

Originally Posted by clayman (Post 1045172)

Here are some more calculations:

Continuing with the same theoretical engines from my previous post:

To rotate one of the engines once, both pistons would have to cycle back-and-forth once. Let's find the total volume of air required to do that: 1.767 cubic inches (fills one chamber of a 4" stroke 3/4" bore piston) x 2 (motions per piston) x 2 (pistons) = 7.068 cubic inches. Multiply that by two again, and you get driving both sides of the drivetrain forward one turn: 14.136 cubic inches.

Remembering that that's in 60 psi, we can half it (this is an educated guess) to get the equivalent in 120 psi, our store pressure: 7.068 cubic inches. If we are using those fancy white plastic tanks (28 cubic inches), that's 3.962 turns of both sides of the drivetrain per tankful. We need some gearing.

Since 4 rotations to a tank is sort of ridiculuously low, lets see how this needs to compare the the 2010 KOP drivetrain (continuing the toughbox thing): In a conservative estimate, the CIM's will be moving at 2000 rpm. With an overall gear reduction of 8.693:1, that works out to the wheels on the robot turning at 230 rpm. With a 2.5 min match, that's 575 wheel rotations in a maximized scenario, 1150 for both wheels.

Now let's figure out how to fit 1150 wheel rotations into, say, 64 tanks. We know that one tank will give us 3.962 engine rotations on both sides per tank, so 64 tanks will give us 254 engine rotations. The gear ratio necessary for each side is easy now: 1:4.53.

Now, once again, back to torque: with the 2010 KOP gear reduction, the wheel torque of those two maxed-out CIM's on one toughbox will be 680 oz-in x 8.693 = 5911 oz-in. With one engine, we have 1600 oz-in. / 4.53 = 353 oz-in. - pretty weak.

You could get this to work out, eventually, but it would require an absolutely ludicrous amount of air tanks. With the 64 tanks in these calculations, you would have to make the robot rather light to get ideal preformance, not to mention that you would spend the entire morning of competition filling those tanks up to 120 psi for one match, at least with anything close to a KOP compressor.

Conclusion: it won't work :rolleyes:

Maybe you should try it with a vacuum engine instead :rolleyes:

You only have to look at the early steam locomotives to realize that a proposal like this is feasable but not practical. Whenever the early engines tried to move at a speed exceeding 5mph (4-cim robot on mecano speed w/ 12.5v), the boiler would be sucked dry, water would have to be put in, pressure would drop, train would stall until more steam was generated. Of course, the efficiency was less than a percent, with the mid-twentieth centry locos pegging 7-8% on a good day.

As for torque: steam/air engines of the piston sort is the only enigne on the planet that can produce maximum force at 0 rpm, unlike gas/diesel. Also, you fellows are probably considering full pressure for the full strike. On a steam engine, the valve gear (monkey motion) was adjusted so that the least amont of steam was needed to do the most work. On startup, steam was admited for maybe 75% of the stroke. As speed increased, the cutoff was reduced so the boiler wouldn't be sucked dry. Volume of steam consumed was decreased, and PV=nRT, so pressure dropped correspondingly. However, more power was produced at higher speeds, with a similar drop in total tractive effort applied to the rail.

For steam enignes, 2-cyl: SP(B^2)(.85)
------------- = TE
D
S = stroke in inches
P = pressure in psi
B = bore in inches
D = driver diametere in inches
TE = tractive effort in pounds (7lbs on clean dry steel rail on 0% slope = 1 ton at constant speed) at START
.85 = accounts for decearse in pressure difference between gauge and piston face
-piston is connected straight to the wheel, no gearbox included
-steam only

New proposal for pneumatic engine: short stroke, large bore, spring return cylinders in pairs pushing against a one way clutch bearing so that the stroke and lever length can be adjusted independently. Pairs are grouped in threes or fours to cover dead-zones created when the cylinders return to their rest positions. Each pair requires its own bearing. The number of groups needed depends on the power requirement, pairs from all groups should be distributed evenly.
for example:

If we use a pair of numbers ,(G,P) ha ha GP, to name the pairs (G would be the group, P would be the pair, and (2,3) would be the 3rd pair of the second group), a simple, 3-speed, 3-pair configuration would look something like this:

(1,1)(2,1)(3,1)[coupler***](1,2)(2,2)(3,2)[coupler***](1,3)(2,3)(3,3)

***to aid in maintenance, the shaft should be split and joined with couplers after every set*.

* a set is composed of the nth pair of all groups
[The first set includes (1,1), (2,1), and (3,1). The first group includes (1,1), (1,2), and (1,3).]

Cylinders of the same pair fire simultaneously, pairs of the same group fire sequentially, and pairs of the same set can either fire simultaneously or sequentially. when cylinders of the same group fire, there should be a 50% overlap for groups of three, and a 33.33% overlap for groups of four, no such overlap is necessary for sets that fire sequentially, but it is acceptable to do so.

Groups are activated in order as more power is needed.

It is also acceptable to create subsets, or groupings of cylinders that fire simultaneously within a set in which all subsets (which must contain the same number of cylinders) fire sequentially.

If subsets are used, lowercase letters following the last parenthesis of the cylinder's name are used to show the subset, with subscripts to show how many groups are active when the cylinder in question is firing, if subscript cannot be used, normal numbers are fine.

For example: (1,1)a4,6; (2,1)b4,6; (3,1)a4c6; (4,1)b4a6; (5,1)b6; and (6,1)c6 make up the first set of an engine with 6 groups that uses subsets when 4 or 6 groups are active.

High grouped arrangements are generally not used, due to weight. The one above is just a hypothetical example used to demonstrate the proper usage of a naming convention.
EDIT: I'm creating a new thread to discuss this topic here. Sorry for off-topic posts.