Question concerning solenoids

R

Robert Scibienski

Guest
Hello,

I am working on a design which will require a solenoid capable
of appx. a one inch throw (is that the correct term?) exerting a force
of appx. one pound with a response time of appx. 0.2 milliseconds. It
will have to fit into a footprint of appx. 1.5 in. sq. Overall length
is of less concern. Are these parameters within the current
capibilities of these devices?
Thanks in advance.
Bob S.
 
A response time of .2 milliseconds is a bit short. This will require a low
impedance coil with ;lots of capacitance in the power supply to drive it
this fast.


"Robert Scibienski" <rjscibienski@sbcglobal.net> wrote in message
news:4154dceb.38875812@news-byoa.prodigy.net...
Hello,

I am working on a design which will require a solenoid capable
of appx. a one inch throw (is that the correct term?) exerting a force
of appx. one pound with a response time of appx. 0.2 milliseconds. It
will have to fit into a footprint of appx. 1.5 in. sq. Overall length
is of less concern. Are these parameters within the current
capibilities of these devices?
Thanks in advance.
Bob S.
Click to expand...
 
On Sat, 25 Sep 2004 03:00:01 UTC, rjscibienski@sbcglobal.net (Robert
Scibienski) wrote:

I am working on a design which will require a solenoid capable
of appx. a one inch throw (is that the correct term?) exerting a force
of appx. one pound with a response time of appx. 0.2 milliseconds.
Click to expand...
I think that requires a force of 415 hp or 300 kW.

IIRC 1 hp = 550 ft lbs, you want to move 1 lb 1 inch in 0.0002 secs. 1
x 0.083 / 0.0002 ~= 415 hp.

Are you sure of your requirements?

--
Jim Backus OS/2 user since 1994
bona fide replies to j <dot> backus <the circle thingy> jita <dot>
demon <dot> co <dot> uk
 
On Sun, 26 Sep 2004 11:37:54 +0000 (UTC), "Jim Backus"
<jhb@nospam.co.uk> wrote:


I think that requires a force of 415 hp or 300 kW.

IIRC 1 hp = 550 ft lbs, you want to move 1 lb 1 inch in 0.0002 secs. 1
x 0.083 / 0.0002 ~= 415 hp.

Are you sure of your requirements?
Click to expand...

I am building a prototype now so that I can measure the exact force
required. The solenoid will be driving a rack to turn a pinion. But
to be honest I never thought of "power" in that way, although I can
see it now. Are you sure this is a correct calculation? I sure hope
not.
Yours in ignorance
Bob S.

ps - is OS2 really worth the effort, or do you just hate MS more than
the average person? <G>
 
"Robert Scibienski" <rjscibienski@sbcglobal.net> wrote in message
news:4157604f.1160968@news-byoa.prodigy.net...
On Sun, 26 Sep 2004 11:37:54 +0000 (UTC), "Jim Backus"
jhb@nospam.co.uk> wrote:


I think that requires a force of 415 hp or 300 kW.

IIRC 1 hp = 550 ft lbs, you want to move 1 lb 1 inch in 0.0002 secs. 1
x 0.083 / 0.0002 ~= 415 hp.

Are you sure of your requirements?


I am building a prototype now so that I can measure the exact force
required. The solenoid will be driving a rack to turn a pinion. But
to be honest I never thought of "power" in that way, although I can
see it now. Are you sure this is a correct calculation? I sure hope
not.
Click to expand...
Well, do you really need 2,500 strokes per SECOND? The noise will
be tremendous, too. This something like 50 times faster than the heads
in a hard disc can seek a track.


--
Graham W http://www.gcw.org.uk/ PGM-FI page updated, Graphics Tutorial
WIMBORNE http://www.wessex-astro-society.freeserve.co.uk/ Wessex
Dorset UK Astro Society's Web pages, Info, Meeting Dates, Sites & Maps
Change 'news' to 'sewn' in my Reply address to avoid my spam filter.
 
On Mon, 27 Sep 2004 03:19:35 +0100, "Graham W"
<graham@his.com.puter.INVALID> wrote:


Well, do you really need 2,500 strokes per SECOND? The noise will
be tremendous, too. This something like 50 times faster than the heads
in a hard disc can seek a track.
Click to expand...

It will not be cycling at that rate, but I need full extension (and
then return following a delay) to occur in that time period if
possible. I have rechecked my calculations and that is what I come up
with - -
Bob S.
 
On Mon, 27 Sep 2004 00:52:05 UTC, rjscibienski@sbcglobal.net (Robert
Scibienski) wrote:

ps - is OS2 really worth the effort, or do you just hate MS more than
the average person? <G
Click to expand...
I've been using 32 bit OS/2 computing since well before Win95 and its
successors appeared and have never seen the benefit in changing (at
least not more my regular computing). Of course I have to use Windows
of various flavours at work and the experience has not encouraged me
to change <G>. It's true that I've never forgiven Bill Gates for
breaking his commitment to support OS/2 - at one time he was quoted as
saying that MS would support OS/2 when it shipped 1 million copies.
When Win95 eventually appeared, the figure for OS/2 shipments was
about 20 million.

--
Jim Backus OS/2 user since 1994
bona fide replies to j <dot> backus <the circle thingy> jita <dot>
demon <dot> co <dot> uk
 
Robert Scibienski wrote:

On Sun, 26 Sep 2004 11:37:54 +0000 (UTC), "Jim Backus"
jhb@nospam.co.uk> wrote:



I think that requires a force of 415 hp or 300 kW.

IIRC 1 hp = 550 ft lbs, you want to move 1 lb 1 inch in 0.0002 secs. 1
x 0.083 / 0.0002 ~= 415 hp.

Are you sure of your requirements?



I am building a prototype now so that I can measure the exact force
required. The solenoid will be driving a rack to turn a pinion. But
to be honest I never thought of "power" in that way, although I can
see it now. Are you sure this is a correct calculation? I sure hope
not.
Yours in ignorance
Bob S.

ps - is OS2 really worth the effort, or do you just hate MS more than
the average person? <G
Click to expand...
(1 inch)*(1 lbf)/(0.0002 sec) = (25 mm)*(4.4N)/(0.0002 sec) = 550 watts.

(1 inch)*(1 lbf)/(0.0002 sec)/(550 ft*lb/hp) = 0.76 Hp.

Does that help? It's still a lot of power.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
"Jim Backus" <jhb@nospam.co.uk> wrote:
It's true that I've never forgiven Bill Gates for
breaking his commitment
Click to expand...
What are you trying to say, Bill Gates lied to us? No Way!
8*|
 
On Mon, 27 Sep 2004 21:15:01 -0700, Tim Wescott
<tim@wescottnospamdesign.com> wrote:

(1 inch)*(1 lbf)/(0.0002 sec) = (25 mm)*(4.4N)/(0.0002 sec) = 550 watts.

(1 inch)*(1 lbf)/(0.0002 sec)/(550 ft*lb/hp) = 0.76 Hp.

Does that help? It's still a lot of power.

Indeed it does help. 500 watts is a lot easier to come by than 300
Click to expand...
kilowatts. Thank you for the reply -
Bob S.
 
On Tue, 28 Sep 2004 19:33:29 GMT, rjscibienski@sbcglobal.net (Robert
Scibienski) wrote:

|On Mon, 27 Sep 2004 21:15:01 -0700, Tim Wescott
|<tim@wescottnospamdesign.com> wrote:
|
|>(1 inch)*(1 lbf)/(0.0002 sec) = (25 mm)*(4.4N)/(0.0002 sec) = 550 watts.
|>
|>(1 inch)*(1 lbf)/(0.0002 sec)/(550 ft*lb/hp) = 0.76 Hp.
|>
|>Does that help? It's still a lot of power.
|>
|Indeed it does help. 500 watts is a lot easier to come by than 300
|kilowatts. Thank you for the reply -
|Bob S.

Yes, but have you considered that to get a solenoid capable of
generating 0.76hp into a footprint of 1.5 in sq is going to take some
remarkable engineering.
 
Ross Herbert wrote:
On Tue, 28 Sep 2004 19:33:29 GMT, rjscibienski@sbcglobal.net (Robert
Scibienski) wrote:

On Mon, 27 Sep 2004 21:15:01 -0700, Tim Wescott
tim@wescottnospamdesign.com> wrote:

(1 inch)*(1 lbf)/(0.0002 sec) = (25 mm)*(4.4N)/(0.0002 sec) = 550
watts.

(1 inch)*(1 lbf)/(0.0002 sec)/(550 ft*lb/hp) = 0.76 Hp.

Does that help? It's still a lot of power.

Indeed it does help. 500 watts is a lot easier to come by than 300
kilowatts. Thank you for the reply -
Bob S.

Yes, but have you considered that to get a solenoid capable of
generating 0.76hp into a footprint of 1.5 in sq is going to take some
remarkable engineering.
Click to expand...
Indeed! And those calcs of poundsf*feet/second etc are the continuous
motion requirement fot lifting a mass against the force of gravity. Like
lifting stones out of a quarry with a crane.

More importantly, no account has been made yet of the acceleration
of the moving parts from the stationary condition to the moving one.
This is likely to exceed the continuous power requirement by quite a lot.

I wish I was a mechanical engineer 'cos this looks like the 300kW
figure may be the correct one!

--
Graham W http://www.gcw.org.uk/ PGM-FI page updated, Graphics Tutorial
WIMBORNE http://www.wessex-astro-society.freeserve.co.uk/ Wessex
Dorset UK Astro Society's Web pages, Info, Meeting Dates, Sites & Maps
Change 'news' to 'sewn' in my Reply address to avoid my spam filter.
 
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