View Full Version : Relation of FT-lbs to pounds of force

RB211

11-17-2010, 10:13 PM

Simple engineering question, I need to push on a dowel pin in a swinging arc motion. (Pitch adjustment on an IVO propeller)

We used a torque wrench and measured that it took 400 ft lbs of force to move.

Now the question is this, we are using a Linear Actuator. The one we want has 400 lbs of force. This dowel pin sits on a .5" arm. Do we need 400 lbs of force from a linear actuator?

J Tiers

11-17-2010, 10:15 PM

The relation for torque is the "lever arm".

0ne foot-lb torque is a pound force acting on a lever a foot long at a right angle to the lever*.

So, a linear actuator with 400lb force must push at right angles to a lever a foot long to exert a torque on the shaft of 400 ft-lb

If your lever were 2 feet long, it would take only 200 lb force. And if it were 6" long, you would need 800 lb force

*

The reason for the "right angle" (90 deg) is that at any other angle, the "effective" lever arm is some other length, not the lever's actual physical length.

Paul Alciatore

11-17-2010, 10:26 PM

You need 400 ft pounds. That means that the product of the force in pounds multiplied by the lever arm length (arc radius) in feet is 400.

So, you must start with one or the other: the force you want to apply or the length of the lever arm. Then you calculate the other.

ForceInPounds = 400 / LeverArmLengthInFeet

OR

LeverArmLengthInFeet = 400 / ForceInPounds

Example 1: Given 1 pound force

LeverArmLengthInFeet = 400 / 1

LeverArmLengthInFeet = 400 feet

Example 2: Given 10 foot lever arm length

ForceInPounds = 400 / 10

ForceInPounds = 40 pounds

All of the above assumes that the force is applied at a right angle to the lever arm. If not, then you must first calculate that component of the force or calculate the actual force needed to generate that normal force from the actual angle of application to the lever arm.

This dowel pin sits on a .5" arm. Do we need 400 lbs of force from a linear actuator?

Does that mean that your lever arm is only .5" from the axis of rotation? If so then you will required 400 times 24 = 9600 lbs of linear force to actuate the mechanism.

Other considerations apply as Jerry took a swipe at. How many degrees of rotation does the arm make from stop to stop? That will make a big difference if the total angular motion is more than about 20 degrees or so.

The Artful Bodger

11-18-2010, 01:12 AM

Simple engineering question, I need to push on a dowel pin in a swinging arc motion. (Pitch adjustment on an IVO propeller)

We used a torque wrench and measured that it took 400 ft lbs of force to move.

Now the question is this, we are using a Linear Actuator. The one we want has 400 lbs of force. This dowel pin sits on a .5" arm. Do we need 400 lbs of force from a linear actuator?

No, if I understand correctly you need 400 ft lbs which you could get with your 400lb actuator by using a 1ft lever, but only while the lever is at 90 degrees from the axis of the actuator.

You could increase the size of the actuator, increase the length of the lever or use some other mechanism to ensure 400lbs force produces 400 ft lbs of torque on your shaft.

Fasttrack

11-18-2010, 01:13 AM

Like Jerry and Evan have pointed out, you have to consider the angle between the force and the lever arm.

Torque is given by the force times the lever arm times the sine of theta. If the force is perpindicular to the lever arm, then it is the formula given by others ( sin(90) = 1 )

T = F*L*Sin(theta)

If the pin swings through a large arc, the torque will drop off quite a bit as the actuator moves (and you run into other obvious linkage problems)

winchman

11-18-2010, 01:45 AM

How did you go about measuring the torque required to "move the dowel pin"?

It sounds like you're working with an adjustable-pitch propellor for a light airplane. The amount of force needed to change the pitch is going to vary quite a bit as the engine speed and other factors change.

Your 400 foot-pounds figure sounds high for static conditions, but I cannot imagine how you used a torque wrench to measure it otherwise.

J Tiers

11-18-2010, 08:44 AM

The constant lever arm is easy to deal with if you can put a pulley or drum or big gear on the shaft. Then you haul on a wire, or move a rack to keep the point of application of force fairly constant.

Even a quadrant or smaller segmental pulley/gear is fine, you don't need any part of the circle that isn't used.

But we have no idea, at least I don't, what the physical configuration is. Perhaps a lack of imagination, but so far the description in words has not translated into a mental picture for me.

The cylinder on the left is the linear actuator motor that mounts on the front of the prop. It has an internal planetary transmission that drives a leadscrew. The leadscrew extends into the space between the prop blade ends and actuates a pair of offset pins that are connected to steel torsion rods that run the length of the prop blades. The ends of the rods are connected to the blades at the tips only. The blades are composite material and are warped to a different pitch as the lever arms are moved by the collar bushing that is driven by the leadscrew.

http://ixian.ca/pics8/ivo.jpg

If it really does take 400 ft lbs of torque to warp the blade, with only a 0.5" lever arm the actuator will need to deliver about 10,000 lbs of thrust.

bob ward

11-18-2010, 10:05 AM

Not sure I understand your set-up RB211. Can you do a Crap-O-CAD sketch?

Thruthefence

11-18-2010, 12:09 PM

I wonder if it will take that much force in actual operation? The aerodynamic forces on the blade (assuming a non-symmetrical airfoil) tends to take it to coarse pitch, and something called a centrifugal twisting moment tends to take it to fine pitch. Certified Props usually have counterweights on the blades, biasing the forces toward fine, (high rpm). So if the pitch change mechanism fails, it goes to high power operation.

Here's a section view of one:

http://www.ivoprop.com/images/inflightpartspic.jpg

The 400 ft lb number sounds high to me. However, a look at a bolt tightening table shows that a torque of 50 ft lbs applied to a 1/2 x 13 tpi bolt with lubrication will provide a clamping force of over 12,000 lbs. That isn't at all inconsistent with the apparent size and design of the actuator motor shown in the picture and the parts list. A motor that size with a large numerical ratio transmission can easily produce 50 ft lbs of torque which then translates to 12,000 lbs of effort with a half inch lead screw. I have some trailer jacks that I use to operate the angle drive on my snowplow blade and the dimensions are very similar. They aren't geared down nearly as much as the planetary gearbox would produce but they are still rated at lifting 5000 lbs each.

RB211

11-18-2010, 02:55 PM

Hello all, let me show you exactly what we have done so far to clear up any questions.

Yes, this is the IVO system we are using. The posted diagrams work perfectly fine for the front prop. Our problem is that we wanted to be different and have a "contra-rotating" power system. This pitch mechanism is for the rear set of propeller blades, hence the need to devise a different pitching mechanism.

How we determined the torque required. Here is the formula straight from the FAA AC 43.13-1B.(A great reference for any machinist)

http://www.flightschoolreview.net/images/torque.jpg

We took a 1" open ended wrench adapter and torque wrench and applied it to the pitching arm of one propeller blade. We had a 20" arm. The torque wrench indicated 240 ft lbs of force for full deflection. Yes, it requires far less force for most of the range.

So.. Y = 240

(X20)/(20)

some fuzzy math...(ie get out my TI92)

X = 240 ft lbs..

Ooh Geez, I guess I really did not need the calculator for that! DOH...

Oh wait... I see an error with units... the 240 is in ft lbs, the length is in inches...

So Y = 2880

(x20)/(20)

So 2880 lb inches... / 12

So 240 ft lbs?

So am I right or did I make myself look like an idiot?

Now for the layout

http://www.flightschoolreview.net/images/rearpitchlayout.jpg

Here is the rest of it...

http://www.flightschoolreview.net/images/twindrive.jpg

RB211

11-18-2010, 02:56 PM

The 400 ft lb number sounds high to me. However, a look at a bolt tightening table shows that a torque of 50 ft lbs applied to a 1/2 x 13 tpi bolt with lubrication will provide a clamping force of over 12,000 lbs. That isn't at all inconsistent with the apparent size and design of the actuator motor shown in the picture and the parts list. A motor that size with a large numerical ratio transmission can easily produce 50 ft lbs of torque which then translates to 12,000 lbs of effort with a half inch lead screw. I have some trailer jacks that I use to operate the angle drive on my snowplow blade and the dimensions are very similar. They aren't geared down nearly as much as the planetary gearbox would produce but they are still rated at lifting 5000 lbs each.

The motor does have a very large gear reduction drive inside of it, along with limit switches.

Ivo does sell 3 and 4 blade setups that use the same motor.

winchman

11-18-2010, 03:05 PM

Here's a picture of a crashed IVO propellor that shows the mechanism used for twisting the blade. Now things are making a little more sense.

http://www.rotaryeng.net/Inside-the-IVO.jpg

You read 240 foot lbs on a 20 inch lever. The correct torque reading cannot be calculated without knowing how long the torque wrench lever arm is without the adapter.

RB211

11-18-2010, 03:33 PM

You read 240 foot lbs on a 20 inch lever. The correct torque reading cannot be calculated without knowing how long the torque wrench lever arm is without the adapter.

I am confused now, the owner did the calculations and re-reading the formula for the purposes of making this post, I think he might of made a mistake. I'll have to ask him about it. For arguments sake however, lets say it IS 240 ft lbs, can my little actuator with 400 lbs of force do the job?

For arguments sake however, lets say it IS 240 ft lbs, can my little actuator with 400 lbs of force do the job?

No. The linear force requirement is somewhere around 6000lbs if the supercams have a 1/2" lever arm from the centre of the spool followers to the centre of the torsion bars. If it is longer than that then divide accordingly. For a 1 inch lever arm on the supercams it is 3000 lbs required.

Thruthefence

11-18-2010, 05:00 PM

Pretty ambitious project, please keep us posted on your project!