I could have come up with it pre-senior. Getting lazy.
Anyway, say it takes 15a at 12vdc to lift 4000lbs 1ft in 90 seconds. How much current would it require to do the same lift in 45 seconds?
Thirty amps? That seems much too simple. :dubious:
But;
12x15x90 = 16200 ws
And;
12x30x45 = 16200 ws
Am I missing something?
Peace,
mangeorge
No you are not missing anything. Lifting the same weight the same distance in half the time requires twice the power. In this case it takes twice the current.
This is assuming, of course, that the motor will still work, and at the same efficiency, with twice the current. But given the problem as stated, you rather have to assume that, since you don’t have anything else to assume.
And it also assumes either a DC system or one with unity power factor if AC.
I was looking to solve for current, which I didn’t state well at all. I don’t remember how to get that element of the formula alone on one side of the “=”.
Ignoring effeciencies and other details, it shouldn’t make any difference how you change time, either through gearing or a larger and faster motor or simply increasing the voltage… The actual work (ws) remains the same.
This whole question came from another board, from a poster who wished his (and many more of us) electric car jack was faster.
It’s actually a pretty cool jack which plugs into your car’s “auxillary power” outlet.
I can post a link if anyone’s interested.
FWIW, gears have a power effeciency, too. This is because gears are actually designed to slip, to maintain a constant angular velocity. The tooth profile must be what’s known as involute, which means it is shaped in such a way that the transfer of power between two meshed gears happens in the same plane. This keeps the radius, and thus the gear ratio, from constantly changing. Due to this slippage, and the coefficient of friction, you’re going to lose power as you start adding more gears.
To move from one side to another, divide both sides by the factors you want to move.
Power = Voltage * Current
P=IV
You’re looking at total energy, (in Watt-Seconds) so multiply by time.
E=IVT
To get current, divide both sides by VT and you get:
I=E/(VT)
If you reduce T by 1/X, you increase current by X.
-Butler
(This is the math you were looking for, right?)
Er, my understanding of gear teeth is that the shape is designed to make the contact between the teeth a rolling contact in order to eliminate as much friction as possibler. The stip occurs because we are unable to generate the proper curve and can only approximate it. Also there is some inherent mismating because we are dividing a whole number of teeth into a curcumference which is an irrational number. Regardles of the change in radius as the point of contact of the teeth moves, when gear A moves one tooth so does gear B. If A has 10 teeth and B 20 then A will rotate twice while B rotates once. The torque on the two gears might fluctuate as the radius changes but the gear ratio is constant. I think.
The above is for spur gears. There are also hypoid gears. For example the set of bevel gears in automotive differentials have the axle of the pinion gear offset of the axle of the ring gear. This is done to lower the drive shaft and eliminate the hump in the floor board. There is a lot of slip in hypoid gears and they are more inefficient that spur gears but the improvement in interior arrangement is worth is. I guess.
Helical (hypoid) gears are also much quieter than spur gears.
Concerning the involute profile of both helical and spur gears, taken from Shigley’s Mechanical Engineering Design book:
“… When the tooth profiles, or cams, are designed so as to produce a constant angular-velocity ratio during meshing, these are said to have conjugate action. In theory, at least, it is possible arbitrarily to select any profile for one tooth and then to find a profile for the meshing tooth that will give conjugate action. One of these solutions is the involute profile, which, with few exceptions, is in universal use for gear teeth and is the only one with which we should be concerned.”
Bolding mine, italics his.
Also, when involute curves are used, “the gears tolerate changes in center-to-center distance with no variation in constant angular-velocity ratio.”
One other thing I found interesting “A pair of gears is really pairs of cams that act through a small arc and, before running off the involute contour, are replaced by another identical pair of cams.”
(bolding mine)
Nit:
The bolded terms are not interchangeable.
Hypoid gears are helically (spiral) cut. But not all spirall cut gears are hypoid. The term hypoid means that the rotational axes of the gears are at right angles, but do not intersect. So spiral cut gears with paralell shafts (typical in automotive transmissons, both automatic and manual) are helical gears, but definatly not hypoid. Neither are all spiral cut bevel gears hypoid. Though they are generally hypoid sets for the last half century or so, there have been holdouts even for vehicular final drives. Toyota land cruisers, and BMW motorcycles spring to mind.
That’s it exactly. Thanks.
Boy, that old adage “use it or lose it” fits, doesn’t it.
Now I’m off to that other board so’s I can look clever. 
mangeorge
Interesting. 
So what are the typical losses in a typical automobile transmission? I’m sure it varies with torque, etc, but there must be an average. I’m thinking the total loss isn’t all that great.
BTW; this subject was addressed in an article I read about the Multitronic (a CVT) tranny in my Audi. The author proposed that less loss in that tranny could be part of the reason it would outperform* a six-speed manual in the same car.
*I know, I know. That’s hard to accept for some.
mangeorge
This is seat-of-my-pants, and I don’t have my textbooks with me, so I’ll look it up tomorrow.
IIRC, .95 is a pretty common assumption for a pair of meshing, steel gears. So, if there are 4 gear reductions in a drivetrain, for example, the efficiency would be .95^4, or about 80%. This is a VERY conservative estimate, and I’d guess an actual transmission would be closer to .99^4, or .96. This still seems like alot of losses, but there’s quite a bit of heat being dissipated, as well as alot of noise. I’ll check in tomorrow with some more info when I’m back with my good friend Shigley.