| Grand Prix Racing - | The Science of Fast Pinewood Cars |
How hard an axle pushes down on its wheel bore determines how much friction will oppose the spinning of that wheel. The effect of each force on your car is discussed and a general expression is derived. In a Grand Prix race, everything is riding on your axles!
When your car is at rest on the track, it bears down on the track with a total force of mg; m is the total mass of your car and g is the acceleration of gravity. But your axles support a combined force that is somewhat less. The weight of the wheels is directly supported by the track so your axles don't carry their weight. The combined force carried by your axles is:
b = m - 2mf - 2mr
b is the mass of your car. It is defined in terms of the total mass, m, the mass of a front and rear wheel, mf and mr in mass units of ozs2/in, ounce-seconds squared per inch (a weight, oz, divided by an acceleration, in/s2).
Each axle supports a part of the body mass depending on how far it is from the center of mass. If the center of mass of your car is in the middle, each wheel supports about one quarter of your car's weight. When the center of mass is closer to the rear, the rear wheels suport more of your car's weight than the front and vice versa.
The car body, assessories and weight of the axles themselves are not the only sources of force (gravitational) on the axles. On tracks that curve downward or upward a psuedo force and rotational force also act. If you could shrink and ride on the axle itself, you would feel a downward force traveling through valleys and an upward one going over hills. You wouldn't feel any extra force on ramps or flat areas. This is called centrifugal force. A small and more subtle force is that created by the rotation of the car body. It is as if the car were put in a hamster wheel with its center of mass in the middle. The force felt would be those needed to spin the wheel to exactly imitate the rotational motion of the car on the track. Through a valley, the front gets a little lift, while the rear is pressed to the track a bit harder. A hill gives the opposite effect. This rotation even gives rise to a little more centrifugal force. But this time it pushes out toward the bumpers!
The total force, a combination of gravitational, centripetal, and rotational force acts through your car's center of mass. Using symbols from the car model and track model, we can find the total force on each axle. We start with the gravitational part of the force, G. Next we look at the centripetal part due to the car's path, C. Rotation is looked at last, first the spinning force, B, then its centrifugal part, P.
The gravitational force, G, acting on each axle can be broken down into the part of force that acts on the front, Gf, and rear, Gr, axles.
G = bg = Gf + Gr since gravity is a conservative force
In order to be in balance around the center of mass, the total angular moment must be zero. The car has a small angular moment as it rotates with the downward or upward bends in the track. But this is not a necessary result of the gravitational force but occurs because of the physical constraint of the track. In order to separate the forces cleanly, zero moment is assumed here. Then,
0 = Gr(CMxcos(O+*)-CMysin(O+*)) - Gf((B-CMx)cos(O+*)+CMysin(O+*)).
O is the angle of the car's foot print to horizontal. The car's foot print is the rectangle between the wheels' contact points on a flat surface. You have to imagine that "rectangle" rotating with the car oon a sloped track. When both wheels are on a ramp then O corresponds to the ramp angle. However, on a downward or upward sloping section of track, the contact angle is different beneath each wheel. In that case, O is typically close to the average of the front and rear wheel contact angles. Note we have defined this "foot print" to be parallel to the car's trajectory. So this "foot print" angle is the same as the trajectory angle at the car's center of mass.
Gravitational force on the front and rear axles can then be solved for using the two expressions above. Find Gf by eliminating Gr, then find Gr by elliminating Gf.
Gf = G(CMxcos(O+*)-CMysin(O+*))/Bcos(O+*)
Front axle: Gf = G(CMx-CMytan(O+*))/B
Gr = G((B-CMx)cos(O+*)+CMysin(O+*))/Bcos(O+*)
Rear axle: Gr = G(B-CMx+CMytan(O+*))/B
Note that when the car has the same size wheels on front and rear and is on level track, the tan(O+*) terms become 0. Then the force felt by the rear wheels is proportional to the distance from the center of mass to the front axle divided by the axle base. The force felt by the front wheels becomes proportional to the distance from the center of mass to the rear axle divided by the axle base.
Centrifugal force squashes your wheels between the track and your car's axles. If it were the only force acting, the axles would touch the wheel bore at an angle at right angles (normal) to the car's foot print (or rather it's trajectory). This force contributes to friction only while the track is curving down or up.
A system of equations is set up to split out the portion of the centrifugal force on the front, Cf, and rear, Cr, axle respectively. The total centrifugal force, C, depends on the mass of the car, b, it's speed, v, and the radius of curvature of its trajectory, R.
C = bv2/R = Cf + Cr
Even though centrifugal force only occurs where the track is curving, the car body rotattion that occurs is not caused by the centrifugal force per sey. So to separate out centrifugal force, assume that there is no angular moment,
0 = Cr(CMxcos(O+*)-CMysin(O+*)) - Cf((B-CMx)cos(O+*)+CMysin(O+*))
Substituting in this with Cr = C - Cf from above, Cf can be isolated.
0 = (C-Cf)(CMxcos(O+*)-CMysin(O+*)) - Cf((B-CMx)cos(O+*)+CMysin(O+*))
Combine terms in Cf,
0 = C(CMxcos(O+*)-CMysin(O+*)) - Cf(CMxcos(O+*)-CMysin(O+*)+(B-CMx)cos(O+*)+CMysin(O+*))
Isolate Cf,
Cf = C(CMxcos(O+*)-CMysin(O+*))/Bcos(O+*)
Simplify,
Cf = C(CMx-CMytan(O+*))/B
Cr is found similarly, eliminating Cf in the moment expression by substituting Cf = G - Cr. The answer is,
Cr = C(B-CMx+CMytan(O+*))/B
Like the gravitational force, if the base slope, *, and foot print angle, O, are zero, the centrifugal force is split by the ratio of distance from the center of mass to the opposite axle.
In order to get the total supporting forces in a form that can be used in the models, it has to be split into parts that lie along the car's trajectory and at right angles to it. The normal force acts at right angles to the car's trajectory. The centrifugal force already acts at right angles to the car's trajectory, but only part of the gravitational force does. The normal force can be written,
N = C + GcosO
Because this normal acts on each axle, it can be distributed to each axle as,
Nf = Cf + GfcosO Normal force on front axle
Nr = Cr + GrcosO Normal force on rear axle
The tangental force acts along the car's trajectory. This time, centrifugal force is not a factor since it only acts at right angles to the car's trajectory.
T = GsinO
Again, we find a part of this force acting on the front axle and part on the rear.
Tf = GfsinO Tangental force on front axle
Tr = GrsinO Tangental force on rear axle
The total force at the car's center of mass due to weight and centrifugal force is the vector sum of the normal and tangent components via Pathagorus' theorem, F = \[N2+T2]. The angle of this force on the axles measured from vertical is q = O - arctan(T/N) (remember O is usually negative so q will be too). Note, when C = 0, q = 0. So, this angle accounts for the centrifugal force which acts normal to the track surface, not straight down like gravity.
Ff = \[Nf2+Tf2], is the force on the front axle.
Fr = \[Nr2+Tr2], is the force on the rear axle.
Substituting for all the intermediate variables we get,
Ff = \[(Cf + GfcosO)2+(GfsinO)2]
Carrying out the square and noting that 1 = sin2O + cos2O,
Ff = \[Cf2 + Gf2 + 2CfGfcosO]
Expanding variables again,
Ff = \[(C(CMx-CMytan*)/B)2 + (G(CMx-CMytan(O+*))/B)2 + 2(C(CMx-CMytan*)/B)(G(CMx-CMytan(O+*))/B)cosO]
Take 1/B2 out of the square root, \[..], and rearrange a bit,
Ff = \[C2(CMx-CMytan*)2 + G2(CMx-CMytan(O+*))2 + 2CGcosO(CMx-CMytan*)(CMx-CMytan(O+*))]/B
Expand the rest of the intermediate variables,
Ff = \[b2v4(CMx-CMytan*)2/R2 + b2g2(CMx-CMytan(O+*))2 + 2b2vgcosO(CMx-CMytan*)(CMx-CMytan(O+*))/R]/B
Similarly, Fr is found from Fr = \[Nr2+Tr2],
Fr = \[(Cr + GrcosO)2+(GrsinO)2]
Fr = \[Cr2 + Gr2 + 2CrGrcosO]
Fr = \[(C(B-CMx+CMytan*)/B)2 + (G(B-CMx+CMytan(O+*))/B)2 + 2(C(B-CMx+CMytan*)/B)(G(B-CMx+CMytan(O+*))/B)cosO]
Fr = \[C2(B-CMx+CMytan*)2 + G2(B-CMx+CMytan(O+*))2 + 2CGcosO(B-CMx+CMytan*)(B-CMx+CMytan(O+*))]/B
Fr = \[b2v4(B-CMx+CMytan*)2/R2 + b2g2(B-CMx+CMytan(O+*))2 + 2b2v2gcosO(B-CMx+CMytan*)(B-CMx+CMytan(O+*))/R]/B
For * = 0, as when the wheels have the same outer radii,
Ff = \[b2v4CMx2/R2 + b2g2(CMx-CMytanO)2 + 2b2v2gCMx(CMxcosO-CMysinO)/R]/B
Fr = \[b2v4(B-CMx)2/R2 + b2g2(B-CMx+CMytanO)2 + 2b2v2gcosO(B-CMx)(B-CMx+CMytanO)/R]/B
For C = 0, as on a ramp,
Ff = bg(CMx-CMytan(O+*))/B
Fr = bg(B-CMx+CMytan(O+*))/B
For * = 0 and C = 0,
Ff = bg(CMx-CMytanO)/B
Fr = bg(B-CMx+CMytanO)/B
The track under the car forces the wheels to follow it up and down beacuse it is rigid and gravity keeps both together. Because the track is rigid, the car can not fall through it. Newton's third law says that every action evokes an equal and opposite reaction. So the track pushes up on the car with the same force. The forces are equal and opposite; they cancel producing tension between the axle and wheel hub, which becomes friction when the wheel spins. This is the normal force, N.
Newton's first law says that a car in straight line motion will continue to go straight unless pushed by a force. When the track bends up and the car follows, the track is exerting extra force on the car to make it follow the track. When the track bends down, less supporting force is exerted. The weight of the car is not fully opposed by the track and the car rotates down. Put simply, the force of the track on the car changes with the shape of the track.
As the track shape changes up or down, the car rotates up or down compared to its trajectory. The force needed to rotate the car is the change in track force that makes the car follow it.
Force acting at right angles to rotate a body is called a torque. Torque on the other hand is also expressed in terms of rotational or angular acceleration, @. In terms of the car model,
I@ = Bf((B-CMx)cos*+CMysin*+Rfsin(Of-O)) = -Br(CMxcos*-CMysin*-Rfsin(Or-O))
The "I" is called the moment of inertia. It basically captures how hard it is to rotate the object.
Using this relation, Bf can be expressed in terms of Br and vice versa. We have to get an idea of what @ is to solve for either. One possible way to do this is to find solutions for Bf and Br based on your car for various Radius of Curvatures. Here the radius of curvature would be an approximation to your car's trajectory and how long that small path is held.
For example, for a 6 foot radius of curvature, held for half a second, use Of = It2 + wt + Or
to get w, the average angular speed during that rotation. Assuming the car started the trajectory with some rotation, w0, the average angular acceleration at that curvature during the half second would have been,
@ = w-w0/t
Then Bf and Br could be estimated as,
Bf = I@/((B-CMx)cos*+CMysin*+Rfsin(Of-O)) and
Br = -I@/(CMxcos*-CMysin*-Rfsin(Or-O))
Strickly speaking, for curved tracks, these forces should be added to Ff and Fr as,
Bf = -Bfcos(Of-O) - Bfsin(Of-O) (normal + tangental part)
Br = -Brcos(Or-O) + Brsin(Or-O) (normal + tangental part)
On the ramp/flat track model, only the weight of the car body acts. In the worst case, the wheel has an infinitely large moment of inertia. This wheel would resist rolling and the axle would settle to the lowest point in the wheel bore, waiting for the wheel to begin rolling.
In the limit of no friction, the axle is content to stay there. When there is friction, it tries to get the axle stuck to the wheel bore and ride backward with it. However, the wheel doesn't want to go as fast as the axle which is driven downward by body weight. It is this push that normally overcomes axle fricition.
If that is not enough push, the axle rides up backward on the wheel bore wall gaining potential energy until it overcomes the friction. This can only happen with a lot of friction.
With even more friction, the axle can "glue" itself to a spot on the wheel bore and travel all the way around the wheel, bouncing the car body up and down. This vibration is the worst enemy your car has. It means your axles are not sliding on the wheel bore, they're rolling on it!
| [Pit Area] | [Title Page] |
| Grand Prix Racing - | The Science of Fast Pinewood Cars |
| Copyright © 1997, 2004 by Michael Lastufka, All rights reserved worldwide. | |