Subject:
|
Re: SSClagorpion - Compressor
|
Newsgroups:
|
lugnet.technic
|
Date:
|
Wed, 2 Jun 2004 15:27:02 GMT
|
Viewed:
|
19642 times
|
| |
|
|
Often, at work on Friday afternoon, we have informal discussions about stuff
completely unrelated to work.
This morning, I stood up and looked over my cube walls (not hard, they're only 4
ft tall) and said I had a question. I didn't get much response, until I said it
was completely unrelated to work. Then, suddenly, everyone was interested...
:)
So, we had a "Friday Afternoon" discussion of compressors, and work/time (NOT
working overtime :)
more below...
In lugnet.technic, Kevin L. Clague wrote:
> I did some measuring and calculating (only outer diameters :^(
>
> A large piston has about 6 times the volume of a small piston. To make a a
> single small pump compete, you would have to turn it over six times as fast as a
> large pump.
I did an experiment, and got the same result. The small pump had to be
compressed six times as much as the large, to provide the same pressure.
> Using six pumps, you could get six small pumps compressing in the
> time it takes to make one large pump compress, but if you did them
> simultaneously, they would provide *more* surface area than the single large
> pump. More surface area at the same pressure means more resitance.
>
> The area of a large piston face is about four times the area of the small
> piston, so six small piston are is larger than a large piston. But......
> assuming you spread the small pumps evenly through a single rotation, then only
> half the pumps are compressing at a time. Also only one of the pistons is
> reaching maximum compression at a time (and at 1/4 of the surface aree of the
> large piston).
So, my general assertion is that the "work" required to compress six small pumps
is the same as the work required to compress one large pump. That should take
into account the "pushback" pressure.
The work required to compress one small pump is 1/4 per unit of distance
(because the area of the piston head is 1/4). Also, the cylinder is
approximately 2/3s the length of the large piston.
Now, I could be wrong (it's been a while sense my last calculus class) but I
think the total work, is the area times the distance: Work=Area*Distance
So, the work required to compress a large piston once would be:
W(l) = A(l) * D(l)
And, for six small piston would be:
W(s) = 6 * A(s) * D(s)
Knowing:
A(s) = 1/4(A(l)) // area of small is 1/4 the area of the large
D(s) = 2/3(D(l)) // small distance traveled is 2/3 large distance
W(s) = 6 * (1/4(A(l)) * (2/3(D(l)) = 6/4 * 2/3 * A(l) * D(l)
W(s) = A(l) * D(l)
and
W(l) = A(l) * D(l)
so
W(l) = W(s)
anyone follow that?
I think the work required for compressing one large piston is the same as the
work required for compressing six small pistons.
Steve
|
|
Message has 1 Reply:
 | | Re: SSClagorpion - Compressor
|
| (...) Ok, so my co-worker (with a degree in physics) said (after many calculations) that the Work is directly related to the Volume. Assuming 6 small pistons pump the same volume as 1 large piston, it will take the same amount of work. Keep in mind, (...) (20 years ago, 2-Jun-04, to lugnet.technic)
|
Message is in Reply To:
| | Re: SSClagorpion - Compressor
|
| I finally had some LEGO time. Given the rate that RC Racer motors suck up current, Using two pumps with two motors and two battery packs, is about the best I can do. I did some measuring and calculating (only outer diameters :^( A large piston has (...) (20 years ago, 1-Jun-04, to lugnet.technic)
|
300 Messages in This Thread:
(Inline display suppressed due to large size. Click Dots below to view.)
- 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
|
|
|
|
