Why Do My Front Wheels Keep Lifting?
Examination of Weight & Loading

ModelWorks 4 inch Scale Burrell Agricultural Traction Engine 

I have written this account primarily for my benefit to document my analysis.  Some of the mathematics may be difficult to follow but as important results are derived careful study will reward the advanced reader with insight into the operation and loading of our miniature tractions engine. If you spot a mistake let me know.

September 2007: This summer I have driven "Little Beastie" over 50 miles and have become very conscious  that it is too light at the front.  At times when I have steered the front wheels and have been embarrassed by engine's refusing to travel in anything but a straight line.  This is particularly noticeable when setting off on soft ground, up hill and will a good load on the trailer.

To aid my studies I have created a mathematical model of the engine.  The starting point to understand my model is the diagram below.

Even this simplified figure is too cluttered and before I could start my analysis I had to make it even more abstract.  For the wheels it is only necessary to consider there contact points with the ground.  Since the front wheel are free to rotate no force other than one perpendicular to the ground are possible and so they are modelled as a frictionless sliding contact.  This is shown in the diagram below as force F which is perpendicular to the ground which slopes at angle q.  The direction that I have shown F may surprise some but it is because the ground pushes up to oppose the weight of the engine. 

The rear wheels are modelled as a pivot whose centre is the ground.  I spent a long time pondering whether to consider the forces at the axle but at the end of the day these are internal forces within the body and and at any moment in time a single point is in contact with the ground about which the weight of the engine pivots.  The frictional forces between wheel and the ground enable force G to be developed which represents the grip of the tyre.

The engine weight W is considered to be a point mass at the centre of gravity of the engine.  I have defined this to be a distance a in front of the rear axle and a distance c above the ground.  The front axle is distance b in front of the rear axle.  The pull on the trailer is represent by force P which is at a perpendicular distance d above the point of contact of the rear wheels with the ground.

While the above text gives the strict definition of my symbols a simplified interpretation is given in the table below.

Now that the notation is established determining the Force on the front wheels is simply obtained by taking moments about the point of contact of the rear wheels with the ground.

bF +dP = [a cos(q) - c sin(q)] W

On the left hand side are the clockwise forces and are fairly simple because b & d are already the perpendicular distances between the line of action of the force and my pivot.  The right hand side considers the anti clock forces and is more complex.  This is because it is necessary to determine the perpendicular distance between the line of action of the force due to the weight of the engine and the pivot.  On the diagram this is the horizontal distance between the extended line of the red arrow of force F and the pivot.  By consider two right angled triangles this can de shown to be  "a cos(q) - c sin(q)".  Rearranging the above equation for F :-

F = a/b cos(q) W - c/b sin(q) W - d/b P

While it may be hard to understand how the above formula was derived it is very illuminating the study the implications of the three terms separately.

Fore - Aft Centre of Gravity
The first term "a/b cos(q) W" accounts for the horizontal position of the centre of gravity.  On level ground q = 0 and since this makes the cos unity the expression becomes a/b * W.  It can be seen that as the centre of gravity is moved from over the rear axle (a=0) to over the front axle (a=b) the load on the front axle varied between 0% and 100% of the engine weight.  This is not at all surprising and is a reminder that moving the C of G  forward would have the beneficial effect  of increasing the front axle load.

I have measured the front axle load on flat ground reported above at 262 lbs.  Assuming the ModelWorks stated weight of the engine at 1100 lbs and a wheels base of 41 enables the horizontal position of the centre of gravity to be calculated. a = 41 * 262 / 1100 = 9.8".  The position of the centre of gravity may also be calculated when the driver is seated on the engine from the 136 lb figure.  This gives a effective engine C of G  4.4" in front of the rear axle when a driver is seated on the bunker seat.  Note the engine weight becomes 1275 as it now has to includes the driver.

The cos(q) multiplier represents the effect of slope of the hill on this terms contribution to the front axle weight.   On a very steep slope of 1 in 7 it works out at 0.99 instead of 1 for flat and therefore the variation of the cos(q) term should be considered insignificant. ie it can be replaced with 1.

Vertical Centre of Gravity
The second term "c/b sin(q) W" accounts for the vertical position of the centre of gravity.  It should be noted that any contribution from this term reduces (minus sign) the front axle load.  Perhaps as expected on flat ground it has no effect as sin(q) will be zero.  The term includes multipliers applied to the total engine weight and since the engine weights 1/2 a ton even a small factor could quickly take many pounds off the front axle. "c/b" represents the height of the centre of gravity of the engine in relation to the wheel base.  

In order to estimate the vertical position of the C of G I repeated the measurement on my sloping drive where I found to front axle load to be 161 lbs.  My drive slopes at 1 in 5 which equates to an angle of 11.3 degrees.  

Back to the formula:-

F = a/b cos(q) W - c/b sin(q) W  

The only "unknown" is c and by substituting F=161, q=11.3 and from above a=9.8, b=41 and W=1100 the vertical position of the centre of gravity is found to be 18.4" giving a value of 0.45 to the c/b multiplier.

Unlike cosine, sine of small angles will have a significant effect on increasing the angle from zero.   Nowadays road signs show hill gradient  as a percentage which is in fact the tangent of the slope angle.  For example  a 45 degree hill  would be 100% and a more gentle 4 degree hill  would be 7% or 1 in 14.  The useful point here is that sin(q) is approximately equal tan(q) for small angles and the value tan(q) is given on the road signs.  Thus we can use this in our formula to estimate the loss of weight from what is written on the road signs.

Loss of weight = c/b sin(q) W
                          = 0.5 * percentage slope/100 * 1275
                          = 638  * percentage slope/100

Since in the case of a seated driver we only have 136 lbs to start we can calculate the angle that will cause the front axle to loose contact with the ground.  Using the version of the formula with the sin as it is slightly more accurate it can be found that the maximum angle is 12.3 degrees or or a slope of 21% (about 1 in 5).

As the angle of slope is increased  from flat to 12.3 degrees the front axle load will proportionately reduce from 136lbs to zero at 12.3 degree. 

Simpler explanation
To provide an alternative viewpoint for those that prefer pictures I have included the graphics on the rights.  The green bulls eye shows the position of the C of G without driver and the red the location with driver sitting on a bunker mounted seat. 

The lower graphic it a repeat of the first but inclined by 12.3 degrees.  Notice that the C of G with seated driver is now vertically above the point of contact with the road.  If the hill were to get any steeper the C of G would be behind the point of contact with the road and the front wheels would lift and continue to lift resulting in .....

It should be noted that this is not the same as lifting the front wheels above the ground whilst the back wheels are on level ground.  Notice that red line showing the vertical extension of the point of contact with the ground is in front of the rear axle. 

Note: These results are for my engine "Little Beastie" with 175 lb driver and while other engines will follow a similar pattern the tipping point will be different.

I recently refused to take "Little Beastie" up a 15% hill for a variety of reasons one of which was I didn't think the front wheels would stay on the ground.  This term shows their would have been a reduction of 96 lbs due to this effect alone from 136 to 40lbs.

The overall front axle load may be restated: -

F = a/b cos(q) W - c/b sin(q) W - d/b sin(q) L

This makes the final term appear and behave much like the nasty middle term.  Thinking of the 15% hill I refused the trailer load would have been about 750lbs making the final terms contribution 36 lbs.  Together with the 96lbs lost thought the raised C of G the flat front axle load of 136 would be reduced by 132 lbs to 4 lbs.  It is now looking as if I would not have made it!

Moving Off and Accelerating
Thus far I have not considered the effect of applying power to accelerate the engine but from experience I know this is when problems occur.  It is well know that force equals mass times acceleration but what accelerations are typical for a miniature traction engine?  I know that a nice steady pace to drive my engine is 3mph and I can reach this in a very quickly.  

The next issue is what units to work in?  The analysis so far has used inches and pounds and  I have decided to continue with imperial units and with express the mass of the engine and trailer in slugs .  By using slugs

Force in pounds=Mass in slugs  * Acceleration in ft per second per second

In order to get an estimate of accelerations I arranged to take videos of of the rear wheels as I accelerated from standstill.  By examining the video frame by fame I was able to determine distance versus time.  Using full throttle in high gear I exceed an acceleration of 6 ft/s/s.  To put that in context an  acceleration of 6 ft/s/s would result in 0 to 60mph in less than 15 seconds.  I suspect the initial acceleration may have been more but with only 30 frames per second I did not have sufficient data until the engine was well under way. 

The mass of the engine in slugs is 1275 / 32 = 40slug and the loaded mass of the trailer 23 slug.  

The force required to accelerate the engine at 6 f/s/s is 40*6=240 lbs and the force required to accelerate the trailer 138 lbs.  It is easy to see how the trailer accelerating force acts on the engine to reduce the front axle load as it is transmitted through the drawbar which is subject to the "d/b P" multiplier. In this case the 138 lbs reduces the front axle load by 44lbs.

It takes a little more though to realise that the 240lbs accelerating the engine is subject to the "c/b" and in this case the 240lbs reduces the front axle load by a whopping 120lbs.

Lightening effect of acceleration = (c/b W/32 + d/b L/32) A 

Where A is the acceleration is feet per second per second.  W & L are in pounds as the /32 factors convert the weights from pounds to slugs.  For a given load and engine the multiplier applied to the acceleration is a constant.  In my case of a seated driver pulling an 750 lb load the weight lost from the front axle will be 27 times the acceleration in feet per second per second.  Recalling that I measured a full throttle acceleration in excess of 6 ft/s/s an so 27*6 = 162 lbs would be lost.

The overall front axle load including acceleration may be restated: -

F = a/b cos(q) W - c/b sin(q) W - d/b sin(q) L - (c/b W/32 + d/b L/32) A 

And with a little more rearranging this become :-

F = a/b cos(q) W - c/b  W (sin(q)+A/32) - d/b L(sin(q)+A/32) 

F = a/b cos(q) W -  (c/b  W + d/b L) (sin(q)+A/32) 

It is interesting to note that acceleration acts in the same way as slope in this equation.  Accelerating at 6/f/s/s on the flat will have the same front axle loading effect as driving at a steady pace up an angle of 10.6 degrees ie a 19% gradient.

Considering "Little Beastie"  with driver seat  on the engine towing a load of 750lbs we can substitute numbers into most of the above equation::-

F =  136 - ( sin(q)+A/32) (0.5 * 1275 + 0.32 * 750)

F =  136 - 877(sin(q)+A/32) 

In order to avoid the front wheels lifting the maximum value of (sin(q)+A/32) is  0.155.  Therefore I could expect the front wheels to be of the point of lifting with an acceleration 5 f/s/s on the flat or when climbing a  8.9 degree slope (15%).  In practice the engine will be accelerated on a slope and neither limits should be approached.  The prediction that the front will lift on the flat is an acceleration of 5 ft/s/s is exceeded agrees with observation as I know that if I use full throttle even when taking off on the flat the wheels will lift.

Soft Ground

The rear wheel rides a "bow wave" moving its contact point forward

This summer has been wet and as a result many rally fields soft.  Often on looking behind I noticed that the back wheels were leaving tracks.  What is not at first obvious is that  this has a significant effect on front axle load.

As the back wheel presses out it furrow it creates a bow wave in front the rear wheel on which the engine rides.  This moves the effective contact point forward from immediately under rear axle towards the bow wave. 

The effect can be most noticeable when the engine has been left for some time on soft ground.  As the engine climbs out of the "pit" in which the back wheel sit the effective contact point could be several inches in front of the rear axle.  With a driver seated on "Little Beastie" the C of G is only 4.4" inches in front of the rear axle and therefore there should be no surprise when the front wheels lift.
 

The solution

Don't Sit on the Engine!

H R Plastow in his writings was adamant that we should not sit on our engines and despite his words the majority of 4" and 4 1/2" owners now sit on their engine.  While I don't totaly agree with his reasoning my equations shows the benefit of riding on the trolley.  If I used a riding trolley they would become: -

F =  262 - ( sin(q)+A/32) (0.45 * 1100 + 0.32 * 925)

F =  262 - 791 (sin(q)+A/32) 

The most beneficial change is the increase in the basic front axle load by 126 lbs from 136 lbs to 262 lbs.  This occurs because the driver is no longer hanging on the back of the bunker pulling the C of G back.  There is also a small improvement in the (sin(q)+A/32) multiplier since the C of G in no longer being raised by the high seated driver.  The absolute operating  limits of the engine with loaded trailer support an acceleration of 10.6 f/s/s on the flat or the ability to climb a  19.3 degree slope (35%) though I doubt that an engine would ever be operated in such conditions.

I have become used to sitting on my engine and was very reluctant to forgo the many advantages so I set about designing a seat that was to all intense and purposes fixed to the engine except that it places the weight of the driver on a fifth wheel under the seat.

My design mounts the seat on a rigid frame under which is mounted a jockey wheel.  The frame is attached to the Burrell by a pair of link arms that allow the frame to move up an down.  Since the link arms are also rigid and are able to resist side loads the seat feels as if it is fixed to the engine.  The ride is very comfortable once the up an down motion of the seat has been got used to.  The engine appears to be a lot happier. This is hard to quantify but it feels less strained and it appears less affected by rough ground.  There is a video of the new seat in action here
 

In practice much less that 180 lbs could be accommodated on the front axle.  A block of lead 12"x3"x2" made to resemble a front axle mounted toolbox would weight around 28 lbs.  This is only tinkering with the problem and can only be expect to offer a marginal improvement.

Having witness the dramatic improvement in engine handling when I stopped sitting on it I am convinced that adding lead is not the solution.  Adding a few pounds is a waste of time so if this route is followed tens of pounds will be needed but bear in mind the additional strain this will place on the engine.

Losing rear tyres
Steve Traill has observed that those unfortunate owners who lose their rear tyres tend to be those who sit on the engine.  Sitting on the engine increases the rear axle load form 826 lbs to 1139 lbs and while I don't believe this is the primary reason for loosing tyres the increase by 1/3 in load will make a poorly bonded tyre more likely to come off.  The increase is nearly twice the driver's weight.  This is because not only does the rear wheel have to take the driver but also the weight removed front the front wheel.