There has been much bandying about on the * *Northwestr CL Forum as of late, regarding how a fuel tank works and what causes an engine to run away. Little of this discussion has actually been based on scientific fact, or even quantifiable observations, so I will in the next few articles try to rectify this problem.

Before I can begin to discuss what happens to the fuel inside a models fuel tank, I will need to introduce the (greatly simplified) topic of motion in a circle, centripetal acceleration, and centripetal force—which we all call—line tension.

Since a detailed study of dynamics is beyond the scope of this website, and completely beyond this article, I am going to set some simplified parameters from this point forward. The plane in question will be flying level, at an altitude equal to the handle, we will assume the lines rake is correct, the plane is pointed tangent to the flight circle, we will disregard the effects of gravity on the lines, the effect they may have on rolling the plane into the circle; and lastly we will assume there are no extraneous aerodynamic forces acting on the plane. Also I will not account for the length of the pilots arm, or any added effect of the circle he/she walks, or any effect of whipping, and back-siding, that may happen. Or to put it differently, we will assume the plane to be a rock on a string.

The measurable parameters we will need to know are: The length of the lines, the mass of the model, and lastly the time to complete a single lap, for the plane. Let us further define the units for each of the aforementioned, to be: Length in feet, mass of the model in slugs (the English unit of mass is the slug and force is in pounds), and lap time in seconds. Centripetal acceleration (A_{c}) is give by the following relationship:

A_{c}=W^{2}r

Where omega (**W**) is the angular velocity given in radians per second, and r is the line length in feet. Before I go into actually making an example calculation, I need to clarify this *Angular Velocity* (A_{c}) concept. There are many ways to measure velocity, the most common and familiar to the reader is, miles per hour (for your car this is average velocity). This is a linear velocity, as it measures linear miles, or it could be feet, as in feet per second. This is not a useful measure for us here, as we have a time for one lap, not for traveling the distance around the circumference of the circle created by the lines. So we will be turning our laps per second into a much more useful measure of Radians per second. Since I digressed to discuss angular velocity, I will now have to double digress to discuss ways of measuring angles. For most people the way to measure an angle is with Degrees (360 in a full circle), or for the civil engineers it is gradient (rise over run). These are not at all useful for our problem because it does not convert linear travel to angular travel, and degree measure has the very pesky unit ^{o} hanging around. So we will be using the most fundamental of angular measure-- in this case, the Radian (Rad). One complete circle in radian measure is defined as: **2****p**. Thinking back to your grade school geometry class, I am sure the reader remembers that the circumference of a circle is given by: **2****p****r**. So all we have here is the same formula without the radius. It also bears mention that there is no unit for the Radian, it is a scalar unit, which makes this exercise possible.

So let’s find the centripetal acceleration of a 48 oz (three pounds) model that has a five (5.00) second lap time, and 70 foot lines. The angular velocity for this model is given by:

W = 2p/5.00 seconds

W = 1.26rad/sec.

Now that we have the angular velocity we need to square it and multiply it by the line length:

A_{c}= (1.26)^{2}x70

A_{c}=111.13 ft/sec^{2} = centripetal acceleration

Now I need to introduce Sir Isaac Newton’s (January 4, 1643 – March 31, 1727) Second Law, force is equal to mass times acceleration:

**F = ma**

In our case the “a” will be A_{c} and the weight as stated earlier will be in pounds and we will have to convert it to slugs.

Now I will need to digress once more to discuss the difference between mass and weight. Mass is a simple measure of how much of a certain substance is present. It is universal, and is not changed by location throughout the known universe. The unit of mass in the metric (S.I.) system is the kilogram, and the unit of force is aptly named the Newton (N), there are 9.8 N (Newtons) per kilogram, because the acceleration due to gravity (here on earth and in average) is 9.8m/sec^{2} (meters per second squared). So one kilogram, times 9.8 meters per second squared, is 9.8 N. In the English system, the unit most of us are familiar with is called the Pound. Pounds (weight) equate to Newtons, and are units of force, not mass, so we will need to convert them to Slugs (the unit of mass in the English system). If you remember the weight of our model was three pounds. Since F=MA we need to find mass so solving for mass gives:

**F/a = m**

The acceleration do to gravity in English units is: 32.4 feet per second squared, so the mass for our model can be given by:

M = 3 pounds / 32.4 ft/sec^{2}

M = .093 slugs.

Keeping Sir Isaac in mind we get from F=MA:

F=.093 slugs x **111.13 ft/sec ^{2}**

F = **10.23 pounds** = Centripetal Force.

Since I have the reader on the edge of their, seat, I think a second example might be of interest. Let’s set some differing parameters and see what happens, with say an 80mph Combat Plane, and then an AMA Class 2 Carrier plane.

__Combat Plane:__ 80mph, 20 ounces in weight, on 60 foot lines:

80mph equates to 3.213 seconds per lap.

2p/3.213=1.96rad/sec = W

A_{c} = W^{2}r, or 1.96^{2} x 60 ft = **229.45 ft/sec ^{2}**

1.25 pounds/32.4 = .039slugs

F = ma

.039 slugs x 229.45 ft/sec^{2} = **8.85 slugs x ft/sec ^{2}** (the units of pounds)

__Class II Carrier Plane:__ 110mph, 56 ounces, in weight, on 60 foot lines:

110mph equates to 2.34 seconds per lap. 2p/2.34 sec=2.69rad/sec = W

A_{c} = W^{2}r, or 2.69^{2} x 60ft = **433.16ft/sec ^{2} **(Holy Cow! That’s over 13 Gs!)

3.5 pounds / 32.4 = .11 slugs

F = ma

.11 slugs x 433.16 ft/sec^{2} = **46.79 pounds** of force!

Now that we have done several calculations regarding line tension, it is interesting to take a look at the rule book for a second. I know you the reader is probably not a rules digit, but bear with me for a second and I will make it worth your while. The Pull test for the events, whose models we have been looking at are as follows: Now that we have done several calculations regarding line tension, it is interesting to take a look at the rule book for a second. I know you the reader is probably not a rules digit, but bear with me for a second and I will make it worth your while. The Pull test for the events, whose models we have been looking at are as follows: Combat 35lbs, Class II Carrier 25G, and Precision Aerobatics—10G. The combat plane pulls 8.85 pounds; while the pull test is 35 pounds or about four times the centripetal force (some call this the safety factor). The models weight was 1.25 pounds, which makes for a 28G pull test.

The Class II Carrier plane has a centripetal force of 46 pounds and a pull test of 87.5 pounds (25 x 3.5lbs) which is a safety factor of less than twice the actual pull of the plane. In terms of G force, the 25G test is also less than twice the centripetal acceleration, so this type of aircraft has what appears to be a very hefty pull but a low safety factor.

The 3 pound Precision Aerobatic model has a 10G pull test of 30 pounds, and in reality 10.32 pounds of centripetal force. This is very close to three times the pull of the model or a safety factor of three.

I am not lobbying for any changes to the rules, I just find it very interesting that the categories of competitive modeling do not have the same multiple of test strength, versus theoretical (calculated) load. Out in the real world engineered safety factors are routinely, 10 times calculated load, and in many cases like bridges they may be 20 times.

Speaking of engineering, now that you can calculate centripetal acceleration, next time you take a cloverleaf freeway off ramp, drive the exact posted speed, and see if you hold the same radius on the road way. If you drift out of the curve, the roadways bank is too low, if you fall in, it is too high. Can you figure out how the engineers know what speed to rate a banked corner? Call it home work but, here is an example for you to try on your own: 100 foot radius curve, posted at 30mph. What angle of inclination is needed for the driver to be able to hold the curve without turning the steering wheel? Hint: It’s the same as those at the big race held Regionals weekend.

Next time, I will be talking about the subject at hand—Gas Tanks!

This page was upated Oct. 19, 2012