flyingagin

OK guys, I was really curious about the line pull given a model weight, speed and line length.
This is not for any particular plane, Just curiosty. These type of math proplems fascinate me.

I have previously tried to calculate line pull before. I really don't think I know what I am doing .

So not being an engineer or having the math abilities myself I went to the following websites to try to figure out line pull.http://www.easycalculation.com/physi...etal-force.php

And used the formula f=mv²/r where f= force in Newton’s and m= mass in kilograms and v=speed in meters per second

I also used the following sites to convert weight and speed to metric and back

http://www.towerhobbies.com/help/convcalcs.html

http://www.unitconversion.org/unit_converter/force.html and selected from Newton [N] to pound-force [lbf]

I then made a spread sheet to convert the weight, speed and radius to metric . Then work the Centripetal Force equation. I then converted back to from metric.

I think I goofed it up.

Anyone know what I did wrong or did I actually get it right?
I cant' upload a .xls type file. It is not an allowed file type. is there a work around?

mikeainia

My Feedback: (1)

In English units, CF in lbs = (W*v^2)/(g*r)
(where W= weight of model in pounds,
v^2= velocity (in ft/sec) squared and ft/sec = mph * (52.8/36) = mph * 1.467
g= gravitational constant (32.14 ft/sec^2)
r = radius of circle in feet (ie: line length)

(W lbs) x ( v ft/sec) x (v ft/sec)
____________________________ . (Note that all the 'ft' & 'sec' cancel out, leaving 'lbs')
. (32.14 ft/sec/sec x r ft) .

So for example = model weighs 40 oz (2.5 lbs), flies on 60 foot lines, at 50 mph
first off: 50 mph is (50 * 1.467) ft/sec = 73.33 ft/sec

so CF = (2.5)*(73.33)*(73.33) / (32.14*60)

= (2.5) * (5377.78) / (1928.4) lb.

= (13444.44) / (1928.4) lb.

= 6.97 lb.

(Will be slightly different, depending on how far you hold your arm out,
how much side force is added or subtracted by rudder offset, engine outthrust, etc.
and also the wind will add or subtract a minor amount depending on whether the
model is upwind or downwind etc.)

Clancy Arnold

Mike
I used your calculation method and substituted the numbers for my CL Scale Taube:
Weight 10Lb 2 Oz. (10.1), Speed 37 MPH(7.5 Sec laps on 65 Ft lines.)
Answer was 14 .24 Lbs line pull. When flying the Taube it does not feel like there iseventhat muchpull on the lines.
Clancy

Jim Thomerson

The usual rule of thumb for stunt models is about three times the weight of the model. Not too far from your calculation.

flyingagin

hahaha
Thanks

Well what I did was only a little off (about 4x to much)never thought of thegravitational constant. I knew my answer was wrong
Thanks I learned a little more today.
Can't work on anything for a few days. Arterial Angiogram thru right wrist. So a 5 lb lift limit on my right hand.

I have had only 1 cl plane that flat out wore me out. Had a piped .65 up front. Flew it on 70 foot lines and simply felt wupped after the motor cut off.

greggles47

Sounds like you need to move the lead outs forward to reduce the line tension.

Obviously not so far that you can't control it.

Regards

G

mikeainia

My Feedback: (1)

The calculation is for the force that is required to keep an object flying at a constant radius. If the object is
a rock, all the force must be supplied by 'pulling' on the object with the line(s).

If the object is a powered aircraft, then some of the force can be supplied by aerodynamic forces produced
by the aircraft itself - lift of the various surfaces and thrust can be used to supply some of that force. Even the
drag of the lines is trying to pull the model into the circle, so that is another bit of force that actually helps the
pilot and is therefore not felt by him. Calculations probably can't be easily done but small variations in the
surfaces can. It should be possible to trim the roll and yaw to supply ALL of the force, such that you feel
no pull at the handle, but you would have to be ready to run when some gust suddenly flings the model into the circle.

flyingagin

Also I found another method (I did not come up with it)
.0668 x lbs x mph squared / line length in feet
this jives with the above method results by mikeainia
I entered the values for Clancy Arnold Scale Taube and got the same results he did.

mikeainia

My Feedback: (1)

That is actually the same equation as I gave with the two constants (1.4666 and 32.14)
pre-calculated into one constant:

(1.466 * 1.466) / 32.14 = .0668

So save yourself some keystrokes with this.

flyingagin

Dang I should have realized that. I kept thinking it looked familar Well I have certainly been known to laugh at myselfe when there wasn't some one else more deservering
I will certainly save the keystrokes I put it in a spread sheet in my control line folder.Thanks againmikeainia

flyingagin

Clancy Arnold

Flyingagin
A good question. I went to the AMA web site and looked at the Membership Manual, page 4,for theSafety Code. The section for CL models is at the bottom. There is no line length statement in the Safety Code. Only that the Control system is to be inspected and given a pull test including the safety thong before flying. This is not all of the CL Safety rules, only the portion affecting Flyingagin's question paraphrazed! That is the rules I would think apply to non competition Control Lineflying.

PLEASE go to the AMA Web Site and read them completelyyourself!!!

As for Competition Flying my Taube is limited to 65 to 70 Foot lines as it weighs over 8 pounds.
Clancy

Dean Pappas

Hello Fyingagain,
As it turns out, the centripetal force calculation is interesting enough, but does not answer the question.
In general, line tension is dictated by aerodynamic trim: rudder offset, line rake, fuselage side area ... in combination with the plane's mass.
Speed ships pull far far less than the centripetal calculation because they are trimmed to fly in a circle just a tiny bit larger than the lines dictate.
If trimmed to fly straight, the drag caused by the outward lift would murder their top speeds.
Meanwhile the Stunt guys are often looking to trim for more than the centripetal force.

best regards,
Dean Pappas

da Rock

Yeah, there is a lot more aerodynamic things that affect the line pull we feel.

Years back, Al Rabe popularized the movable rudder for CL aerobatics airplanes. It moved BOTH WAYS. Yes, it actually yawed the model into and out of the circle. It worked both ways too.

There are gyroscopic precession and p-effect, a couple of aerodynamic things that affect airplanes that are pitching up or down, that affect models too. They're airplanes after all.

His rudder moved with the other surfaces to increase tension when the aero was decreasing and to decrease when it was increasing.

Just mentioned it because it shows how sensitive our CL models are to the aero effects and the trim of the model.

ferocious

With that big a plane be glad it may not be pulling full strength. For a heavy scale plane it is usual to trim it so it is actually fly into the circle a little bit to reduce the CF. Maybe trim it a bit tail heavy, use a bit of left thrust, reduce tip weight so it just accelerates straight ahead on take off, move the leadouts forward a bit, maybe bank the plane to the left a few degrees. It doesn't take much trim like this to cut the line pull significantly. It would be a dangerous stunt plane but can fly just fine for scale and even do a few loops and wingovers if you are careful about the wind.

A 10 lb plane and 14 lb. of line pull is getting to where you should be using .021 lines, just in case the plane gets loose on the lines and heads across the circle.

paw080

Thanks for the math exerciser. I ran my CAT-JET(old Hoyt jet kit) specs through your formula; 155mph,

2lb flying wgt, 70'x .018 solids. The model pulls 46lbs in flight at the speed. I feel very sorry for the serious

jet flyers that play at 210-220 mph.

Tony