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Author Topic: Question in probably the wrong forum, but...  (Read 6991 times)

Offline beedees

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Question in probably the wrong forum, but...
« on: June 27, 2007, 06:37:18 PM »
If I have a shaft with 4 arms attached to it at 90 degree spacing each 1 foot long and a force of 10 pounds is applied to the end of each arm, how much force is being applied to the shaft? 10 pounds, 40 0r somewhere in between? I'm thinking 10. ???

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Offline IronHead

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Re: Question in probably the wrong forum, but...
« Reply #1 on: June 27, 2007, 06:46:35 PM »
you have 0 foot pounds of torque , because you have balance . If you have one 12" lever with 10 pounds at the end ,you have  10 foot pounds of torque.

Offline beedees

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Re: Question in probably the wrong forum, but...
« Reply #2 on: June 27, 2007, 07:23:58 PM »
you have 0 foot pounds of torque , because you have balance . If you have one 12" lever with 10 pounds at the end ,you have  10 foot pounds of torque.
Why do wind turbines, etc. have more than one  blade, then...or two at most? What I didn't put in first post is that all the force is applied in the same dirction like a turbine .

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Re: Question in probably the wrong forum, but...
« Reply #2 on: June 27, 2007, 07:23:58 PM »
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Offline innovation_station

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Re: Question in probably the wrong forum, but...
« Reply #3 on: June 27, 2007, 07:58:08 PM »
will the torque not increase with the speed of what ever you are turning ?


is

Offline IronHead

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Re: Question in probably the wrong forum, but...
« Reply #4 on: June 27, 2007, 08:10:47 PM »
I see, then it is attached to an applied force. Now you getting into inertia and centrifugal force . I can't really answer this one . One of the math guys around here might answer this.

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Re: Question in probably the wrong forum, but...
« Reply #4 on: June 27, 2007, 08:10:47 PM »
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Offline fleebell

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Re: Question in probably the wrong forum, but...
« Reply #5 on: June 27, 2007, 08:41:28 PM »
If you have 10 pounds of force on each arm continious pushing in the same direction you have 40 pounds of torque on the shaft. The output of the shaft depends on the speed that it's turning. 
 A real simple formula that will give you a rough ball park figure is:
     foot pounds torque X rpm / 5252 = max possible horsepower   

 This does not take in to account system friction or any other losses that would have to be subtracted from the total.



Offline beedees

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Re: Question in probably the wrong forum, but...
« Reply #6 on: June 27, 2007, 08:50:55 PM »
you have 0 foot pounds of torque , because you have balance . If you have one 12" lever with 10 pounds at the end ,you have  10 foot pounds of torque.
Why do wind turbines, etc. have more than one  blade, then...or two at most? What I didn't put in first post is that all the force is applied in the same dirction like a turbine .
Never mind, IronHead, being the stubborn a** I am,.....went to the shop and built a simple jig The other 3 are redundant . Net torque increase =zip, zero,,,nada. I can pretty well understand things in the concrete....just have trouble with the abstract. Still just have the orig.10 pounds.
     

#2 No, the torque applied is a set value. ;D ;D
« Last Edit: June 27, 2007, 09:52:35 PM by beedees »

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Re: Question in probably the wrong forum, but...
« Reply #6 on: June 27, 2007, 08:50:55 PM »
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Offline Low-Q

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Re: Question in probably the wrong forum, but...
« Reply #7 on: June 27, 2007, 11:19:59 PM »
If I have a shaft with 4 arms attached to it at 90 degree spacing each 1 foot long and a force of 10 pounds is applied to the end of each arm, how much force is being applied to the shaft? 10 pounds, 40 0r somewhere in between? I'm thinking 10. ???
Torque is in relationship with the distance to the shafts center. If the shaft is 1 inch in diameter, the torque at the shafts surface will be approx 12inch (1 foot) times 10 pounds times 4 = 480 pounds. There is limitless torque in the very center of the shaft.

Br.

Vidar

Offline beedees

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Re: Question in probably the wrong forum, but...
« Reply #8 on: June 28, 2007, 12:07:12 AM »
Be very nice if you and fleebell were right, but I don't think so. ;D ;D

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Re: Question in probably the wrong forum, but...
« Reply #8 on: June 28, 2007, 12:07:12 AM »
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Offline fleebell

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Re: Question in probably the wrong forum, but...
« Reply #9 on: July 20, 2007, 04:30:54 AM »
four arms with 10 lbs each or 1 arm with 40lb on a 1 ft arm  still equals 40 ft lbs on the shaft.  that is a static figure.... once they start moving it will be a different ballgame altogether.

Lee B
www.packratworkshop.com


Offline Dingus Mungus

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Re: Question in probably the wrong forum, but...
« Reply #10 on: July 20, 2007, 06:16:11 AM »
480 lb-in of torque. ;)

~Dingus Mungus


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Re: Question in probably the wrong forum, but...
« Reply #10 on: July 20, 2007, 06:16:11 AM »
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Offline fleebell

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Re: Question in probably the wrong forum, but...
« Reply #11 on: July 20, 2007, 06:29:02 AM »
which equals 40 ft lbs like I said.    in lbs * .0833 = ft lbs

Offline Dingus Mungus

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Re: Question in probably the wrong forum, but...
« Reply #12 on: July 20, 2007, 06:37:38 AM »
which equals 40 ft lbs like I said.    in lbs * .0833 = ft lbs


No thats the equation to convert foot pounds to inch pounds...
http://www.google.com/search?hl=en&safe=off&q=pounds+per+inch+to+pounds+per+foot

Try again...

Offline fleebell

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Re: Question in probably the wrong forum, but...
« Reply #13 on: July 20, 2007, 07:14:12 AM »
No,  I suggest you do because you still have it backwards. Try a real source of info like a physics manual like I did instead of google.    But if you must use the internet as a source try this one as it's at least accurate. The question stated the arms were 1 foot long.    40 ft lb = 480 in lbs  same thing!

http://en.wikipedia.org/wiki/Foot-pound_force   1 ft. lb = 12 in. lbs

Offline Dingus Mungus

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Re: Question in probably the wrong forum, but...
« Reply #14 on: July 20, 2007, 08:57:03 AM »
DOH! I was thinking if the motor is reciving a total of 480 pounds @ one inch from the axle, then you would take your one inch of force and times it by 12 to get your foot pounds of force, but I guess I did have it totally backwards! Kinda counter intuitive.

Conversions
1 foot-pound is equivalent to:

1.355 817 948 331 400 4 joule (J) (exactly)
13 558 179.483 314 004 ergs (erg) (exactly)
~0.001 285 067 British Thermal Unit IT (BtuIT or BTUIT)
~0.323 832 calorie IT (calIT) or ~0.000 323 832 "food calorie" (kcal or Cal)
~32.174 049 foot-poundals (ft pdl)
12 inch-pound force (in?lbf or sometimes informally written: ″#)
192 inch-ounce force (in?ozf)
1.3558179 N?m

 :-[
:: walk of shame ::
~Dingus Mungus

 

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