Appendix 1 | Kudlich-Slibar
Introduction
Collision physics in Virtual CRASH is based on rigid body dynamics. By default, vehicle versus vehicle impacts in Virtual CRASH are simulated using the Kudlich-Slibar impulse-momentum model. This model dates back to the 1960s [1, 2] and is the basis for other vehicle collision simulators used for accident reconstruction [3, 4]. Impulse-momentum simulators are based on Newton's Laws. Note, similar impulse-momentum calculations are commonly performed by accident reconstructionists, often by hand, or using other standard calculation applications [5]. The benefit of calculations based on conservation laws of physics is that they bypass the need to model the specific details of how collision forces dynamically evolve, which reduces the problem of predicting collision outcomes to a problem of accounting for known conserved quantities – namely, momentum and energy. Below is a derivation illustrating the Kudlich-Slibar model.
Velocity at Arbitrary Point within a Rigid Body
In order to derive the standard equations needed to describe rigid body dynamics, we first start by defining the total velocity at any given point \(P\) on or within an extended object \(k\). We obtain this by first noting the position of \(P\) can be expressed as the vector sum:
$$\bar{R}_{k}^{P} = \bar{R}_{k}^{CG} + \bar{r}_{k}^{P} \tag{1}$$
where \(\bar{R}_{k}^{P}\) is the Earth-frame location of \(P\), \(\bar{R}_{k}^{CG}\) is the object k center-of-gravity (CG), and \(\bar{r}_{k}^{P}\) is the location of \(P\) measured with respect to the object’s CG. Let us now assume that while point \(P\) may be free to rotate with respect to the CG, the magnitude of \(\bar{r}_{k}^{P}\) remains constant with time. In this case, taking the time derivative of both sides above, we obtain the expression:
$$\bar{v}_{k}^{P} = \bar{v}_{k}^{CG} + \bar{\omega}_{k} \times \bar{r}_{k} \tag{2}$$
where \(\bar{v}_{k}^{CG}\) is the velocity at the center-of-gravity of object k. \(\bar{\omega}_{k}\) is the angular velocity of object k measured about its center-of-gravity, and \(\bar{r}_{k}\) is the position vector extending from object \(k\)’s CG to point \(P\).
Newton’s 2nd Law
We can express Newton’s 2nd Law in terms of its consequences to the momentum of an object. That is, expressing the momentum of object \(k\) as the product of its mass and velocity:
$$\bar{p}_{k} = m_{k} \bar{v}_{k} \tag{3}$$
we have the total force on object \(k\) given by:
$$\bar{F}_{k} = \frac{d}{dt} (\bar{p}_{k}) = \frac{d}{dt} (m_{k} \bar{v}_{k}) \tag{4}$$
Newton’s 3rd Law
From Newton’s 3rd Law, we know that the forces exerted by two objects on one another are equal in magnitude and opposite in direction. That is:
$$\bar{F}_{12} = -\bar{F}_{21} \tag{5}$$
where \(\bar{F}_{12}\) is the force on object 1 due to its interaction with object 2, and \(\bar{F}_{21}\) is the force on object 2 due to its interaction with object 1.
Let us now express the total force on object 1 as the sum over the interaction force with object 2 and any other external forces:
$$\bar{F}_{1} = \frac{d}{dt} (m_{1} \bar{v}_{1}) = \bar{F}_{12} + \sum_{i=0}^{N} \bar{f}_{1,ext}^{i} \tag{6}$$
where \(\bar{f}_{1,ext}^{i}\) are external forces acting on object 1 not attributable to object 2. Similarly for object 2 we have:
$$\bar{F}_{2} = \frac{d}{dt} (m_{2} \bar{v}_{2}) = -\bar{F}_{12} + \sum_{i=0}^{M} \bar{f}_{2,ext}^{i} \tag{7}$$
Conservation of Linear Momentum
Let us now suppose that during the duration of interaction between objects 1 and 2, the magnitude of the interaction forces greatly exceeds the total external forces on each object, such as during motor vehicle collisions; in this case, we have:
$$\bar{F}_{1} = \frac{d}{dt} (m_{1} \bar{v}_{1}) \approx \bar{F}_{12} \tag{8}$$
and
$$\bar{F}_{2} = \frac{d}{dt} (m_{2} \bar{v}_{2}) \approx -\bar{F}_{12} \tag{9}$$
Summing the total force on the system, we have:
$$ \bar{F}_{1} + \bar{F}_{2} = \frac{d}{dt} (m_{1} \bar{v}_{1}) + \frac{d}{dt} (m_{2} \bar{v}_{2}) $$
$$= \frac{d}{dt} (m_{1} \bar{v}_{1} + m_{2} \bar{v}_{2}) $$
$$= \frac{d}{dt} (\bar{p}_{1} + \bar{p}_{2}) $$
$$\approx \bar{F}_{12} - \bar{F}_{12} = 0 \tag{10} $$
or
$$\bar{p}_{1} + \bar{p}_{2} \approx \text{constant} \tag{11}$$
This expression, of course, tells us that the sum of the linear momentum associated with objects 1 and 2 is conserved over the time in which external forces can be reasonably neglected. This is a typical approximation used for vehicle collisions, which have very brief collision pulse widths, with large associated contact forces exchanged between vehicles. Note, in the absence of any net external forces equations (10) and (11) are exactly true. We will neglect the role of external forces in the rest of this treatment.
We can now write an expression relating the pre-impact to post-impact sum over momenta. This is given by:
$$\bar{p}_{1,i} + \bar{p}_{2,i} = \bar{p}_{1,f} + \bar{p}_{2,f} \tag{12}$$
or
$$m_{1} \bar{v}_{1,i} + m_{2} \bar{v}_{2,i} = m_{1} \bar{v}_{1,f} + m_{2} \bar{v}_{2,f} \tag{13}$$
Where the subscript \(i\) indicates that the values are obtained immediately prior to impact, and \(f\) indicates the values are obtained immediately after impact. We can now rewrite equation (13) as:
$$m_{1} \cdot (\bar{v}_{1,f} - \bar{v}_{1,i}) = -m_{2} \cdot (\bar{v}_{2,f} - \bar{v}_{2,i}) \tag{14}$$
or
$$m_{1} \cdot \Delta \bar{v}_{1} = -m_{2} \cdot \Delta \bar{v}_{2} \tag{15}$$
Impulse
Let us define the total impulse imparted to object \(k\) by:
$$\bar{J}_{k} = \int_{0}^{\Delta t} dt \cdot \bar{F}_{k} \tag{16}$$
where \(\bar{F}_{k}\) is the total force acting on object \(k\) during the duration \(\Delta t\). Using (4) we can rewrite this as:
$$\bar{J}_{k} = \int_{0}^{\Delta t} dt \cdot \left[ \frac{d}{dt} (m_{k} \bar{v}_{k}) \right] \tag{17}$$
which is simply:
$$\bar{J}_{k} = m_{k} \cdot (\bar{v}_{k}(t=\Delta t) - \bar{v}_{k}(t=0))$$
$$= m_{k} \cdot \Delta \bar{v}_{k} = \Delta \bar{p}_{k} \tag{18}$$
or
$$\bar{p}_{k,f} = \bar{p}_{k,i} + \bar{J}_{k} \tag{19}$$
This therefore implies:
$$\bar{v}_{k,f} = \bar{v}_{k,i} + \frac{\bar{J}_{k}}{m_{k}} \tag{20}$$
This is an expression of the impulse-momentum theorem.
Let us again assume that during the interaction period, the forces on each object are dominated by their mutual contact interactions. In this case we have:
$$\bar{J}_{1} = \int_{0}^{\Delta t} dt \cdot \bar{F}_{12} \tag{21}$$
and
$$\bar{J}_{2} = \int_{0}^{\Delta t} dt \cdot \bar{F}_{21} = -\int_{0}^{\Delta t} dt \cdot \bar{F}_{12} \tag{22}$$
and therefore:
$$\bar{J}_{1} = -\bar{J}_{2} \tag{23}$$
Torque
For illustrative purposes, we will derive the impulse relationships for the special case of rotation about the local \(\hat{z}\) axes of our objects which we take as aligned with the global \(\hat{z}\) axis. We can write an expression for the total torque on object \(k\) caused by the application of force \(\bar{F}_{k}\):
$$\bar{\Gamma}_{k} = I_{k} \bar{\alpha}_{k} = \bar{r}_{k} \times \bar{F}_{k} \tag{24}$$
where \(\bar{\alpha}_{k} = \frac{d\bar{\omega}_{k}}{dt}\) is the angular acceleration about the center-of-gravity of object \(k\), and \(\bar{r}_{k}\) is the lever-arm extending from the center-of-gravity to the point of contact, and \(I_{k}\) is the moment-of-inertia for rotation about the local \(\hat{z}\) axis.
Taking the time integral of the total torque over interaction duration \(\Delta t\), we have:
$$\int_{0}^{\Delta t} dt \cdot \bar{\Gamma}_{k} = \int_{0}^{\Delta t} dt \cdot \left(I_{k} \bar{\alpha}_{k} \right)$$
$$= \int_{0}^{\Delta t} dt \cdot \left[ I_{k} \left( \frac{d\bar{\omega}_{k}}{dt} \right) \right]$$
$$= I_{k} \Delta \bar{\omega}_{k} = \Delta \bar{L}_{k} \tag{25}$$
Therefore the torque delivered over time \(\Delta t\) is associated with a change in angular momentum \(\Delta \bar{L}_{k}\), where the angular momentum is given by \(\bar{L}_{k} = I_{k} \bar{\omega}_{k}\).
Therefore from (18) and (25), we have:
$$\Delta \bar{L}_{k} = I_{k} \Delta \bar{\omega}_{k} = \int_{0}^{\Delta t} dt \cdot \left(\bar{r}_{k} \times \bar{F}_{k} \right)$$
$$= \bar{r}_{k} \times \int_{0}^{\Delta t} dt \cdot \bar{F}_{k}$$
$$= \bar{r}_{k} \times \Delta \bar{p}_{k}$$
$$= \bar{r}_{k} \times \bar{J}_{k} \tag{26}$$
From this, we have:
$$\bar{\omega}_{f} = \bar{\omega}_{i} + \frac{\bar{r}_{k} \times \Delta \bar{p}_{k}}{I_{k}}$$
$$= \bar{\omega}_{i} + \frac{\bar{r}_{k} \times \bar{J}_{k}}{I_{k}} \tag{27}$$
Returning now to equation (2), we have for the post-impact velocity at the point of contact \(P\):
$$\bar{v}_{k,f}^{P} = \bar{v}_{k,f}^{CG} + \bar{\omega}_{k,f} \times \bar{r}_{k} \tag{28}$$
Using (20) and (27), (28) becomes:
$$\bar{v}_{k,f}^{P} = \left(\bar{v}_{k,i} + \frac{\bar{J}_{k}}{m_{k}}\right) + \left(\bar{\omega}_{i} + \frac{\bar{r}_{k} \times \Delta \bar{p}_{k}}{I_{k}}\right) \times \bar{r}_{k}$$
$$= \left(\bar{v}_{k,i} + \bar{\omega}_{i} \times \bar{r}_{k}\right) + \frac{\bar{J}_{k}}{m_{k}} + \left(\frac{\bar{r}_{k} \times \bar{J}_{k}}{I_{k}}\right) \times \bar{r}_{k}$$
$$= \bar{v}_{k,i}^{P} + \frac{\bar{J}_{k}}{m_{k}} + \left(\frac{\bar{r}_{k} \times \bar{J}_{k} \times \bar{r}_{k}}{I_{k}}\right) \tag{29}$$
where we use the triple vector product which we write as \(\bar{r}_{k} \times \bar{J}_{k} \times \bar{r}_{k}\). This is given by:
$$\bar{r}_{k} \times \bar{J}_{k} \times \bar{r}_{k} \equiv (\bar{r}_{k} \times \bar{J}_{k}) \times \bar{r}_{k} = \bar{r}_{k} \times (\bar{J}_{k} \times \bar{r}_{k})$$
$$= \bar{J}_{k} \left(|\bar{r}_{k}|^{2}\right) - \bar{r}_{k} \left(\bar{r}_{k} \cdot \bar{J}_{k}\right) \tag{30}$$
Therefore for object \(1\), we have:
$$\bar{v}_{1,f}^{P} = \bar{v}_{1,i}^{P} + \frac{\bar{J}_{1}}{m_{1}} + \left(\frac{\bar{r}_{1} \times \bar{J}_{1} \times \bar{r}_{1}}{I_{1}}\right) \tag{31}$$
Similarly for object \(2\), we have:
$$\bar{v}_{2,f}^{P} = \bar{v}_{2,i}^{P} + \frac{\bar{J}_{2}}{m_{2}} + \left(\frac{\bar{r}_{2} \times \bar{J}_{2} \times \bar{r}_{2}}{I_{2}}\right) \tag{32}$$
Taking the difference between equations (31) and (32), and using (27), we have the final relative velocity vector (separation velocity) at point \(P\):
$$\bar{v}_{\text{Rel,f}}^{P} = \bar{v}_{1,f}^{P} - \bar{v}_{2,f}^{P}$$
$$= \left[\bar{v}_{1,i}^{P} + \frac{\bar{J}_{1}}{m_{1}} + \left(\frac{\bar{r}_{1} \times \bar{J}_{1} \times \bar{r}_{1}}{I_{1}}\right)\right] - \left[\bar{v}_{2,i}^{P} + \frac{\bar{J}_{2}}{m_{2}} + \left(\frac{\bar{r}_{2} \times \bar{J}_{2} \times \bar{r}_{2}}{I_{2}}\right)\right]$$
$$= \left(\bar{v}_{1,i}^{P} - \bar{v}_{2,i}^{P}\right) + \bar{J}_{1} \left(\frac{1}{m_{1}} + \frac{1}{m_{2}}\right) + \left(\frac{\bar{r}_{1} \times \bar{J}_{1} \times \bar{r}_{1}}{I_{1}} + \frac{\bar{r}_{2} \times \bar{J}_{1} \times \bar{r}_{2}}{I_{2}}\right)$$
$$= \bar{v}_{\text{Rel,i}}^{P} + \frac{\bar{J}_{1}}{\bar{m}} + \left(\frac{\bar{r}_{1} \times \bar{J}_{1} \times \bar{r}_{1}}{I_{1}} + \frac{\bar{r}_{2} \times \bar{J}_{1} \times \bar{r}_{2}}{I_{2}}\right) \tag{33}$$
where we use the reduced mass given by:
$$\bar{m} = \left(\frac{1}{m_{1}} + \frac{1}{m_{2}}\right)^{-1} \tag{34}$$
Let us suppose we can model the impulse on object as being caused by effects along the normal axis, \(\hat{n}\), generally associated with the direction along which material compresses during contact on object 1, and the axis tangent to \(\hat{n}\), \(\hat{t}\), which is given by:
$$\hat{t} = \frac{\left(\hat{v}_{\text{Rel,i}}^{P} \times \hat{n}\right) \times \hat{n}}{\left|\left(\hat{v}_{\text{Rel,i}}^{P} \times \hat{n}\right) \times \hat{n}\right|} \tag{35}$$
Thus, \(\hat{t}\), is contained in the plane orthogonal to \(\hat{n}\), and whose direction is fixed to point antiparallel to the projection of the closing-velocity vector on this plane. Here let us also define the axis: \(\hat{z}' = \hat{t} \times \hat{n}\). The impulse component along \(\hat{t}\) is generally associated with frictional effects, and is related to the impulse component along \(\hat{n}\) by:
$$\mu = \frac{\bar{J} \cdot \hat{t}}{\bar{J} \cdot \hat{n}} \tag{36}$$
where \(\mu\) is identified with the coefficient-of-friction between the two objects undergoing contact interaction.
Making a connection to other known planar impact mechanics presentations, let us define the coefficient-of-restitution as:
$$\varepsilon = -\frac{\bar{v}_{\text{Rel,f}}^{P} \cdot \hat{n}}{\bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n}} \tag{37}$$
With this definition, we can now rewrite (33) after projecting both sides onto the normal axis \(\hat{n}\):
$$-(1+\varepsilon) \cdot \bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n} = \frac{\bar{J}_{1} \cdot \hat{n}}{\bar{m}} + \left(\frac{\bar{r}_{1} \times \bar{J}_{1} \times \bar{r}_{1}}{I_{1}} + \frac{\bar{r}_{2} \times \bar{J}_{1} \times \bar{r}_{2}}{I_{2}}\right) \cdot \hat{n}$$
$$= |\bar{J}_{1}| \left(\frac{\hat{J}_{1}}{\bar{m}} + \frac{\bar{r}_{1} \times \hat{J}_{1} \times \bar{r}_{1}}{I_{1}} + \frac{\bar{r}_{2} \times \hat{J}_{1} \times \bar{r}_{2}}{I_{2}}\right) \cdot \hat{n} \tag{38}$$
We can finally solve for the magnitude of the impulse on object 1 by:
$$|\bar{J}_{1}| = \frac{-(1+\varepsilon) \cdot \bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n}}{\left(\frac{\hat{J}_{1}}{\bar{m}} + \frac{\bar{r}_{1} \times \hat{J}_{1} \times \bar{r}_{1}}{I_{1}} + \frac{\bar{r}_{2} \times \hat{J}_{1} \times \bar{r}_{2}}{I_{2}}\right) \cdot \hat{n}} \tag{39}$$
Let us now fix the orientation of the normal axis to point parallel with the normal component of the impulse on object 1, and antiparallel with the normal component of the impulse on object 2.
We can now express the impulse on object 1 in the vector form:
$$\bar{J}_{1} = \frac{|\bar{J}_{1}| \left(\hat{n} + \mu \hat{t}\right)}{\sqrt{1+\mu^{2}}} \tag{40}$$
The unit vector \(\hat{J}_{1}\) is given by:
$$\hat{J}_{1} = \frac{\left(\hat{n} + \mu \hat{t}\right)}{\sqrt{1+\mu^{2}}} \tag{41}$$
where
$$\hat{J}_{1} \cdot \hat{n} = \frac{1}{\sqrt{1+\mu^{2}}} \tag{42}$$
and
$$\hat{J}_{1} \cdot \hat{t} = \frac{\mu}{\sqrt{1+\mu^{2}}} \tag{43}$$
Again, the triple vector products can be expressed by:
$$\bar{r}_{1} \times \hat{J}_{1} \times \bar{r}_{1} = \hat{J}_{1} \left(|\bar{r}_{1}|^{2}\right) - \bar{r}_{1} \left(\bar{r}_{1} \cdot \hat{J}_{1}\right) \tag{44}$$
and
$$\bar{r}_{2} \times \hat{J}_{1} \times \bar{r}_{2} = \hat{J}_{1} \left(|\bar{r}_{2}|^{2}\right) - \bar{r}_{2} \left(\bar{r}_{2} \cdot \hat{J}_{1}\right) \tag{45}$$
Therefore, using \(42\), \(44\), and \(45\), we can rewrite \(39\) as:
$$|\bar{J}_{1}| = \frac{-(1+\varepsilon) \cdot \bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n}}{\left(\frac{\hat{J}_{1}}{\bar{m}} + \frac{\hat{J}_{1} \left(|\bar{r}_{1}|^{2}\right) - \bar{r}_{1} \left(\bar{r}_{1} \cdot \hat{J}_{1}\right)}{I_{1}} + \frac{\hat{J}_{1} \left(|\bar{r}_{2}|^{2}\right) - \bar{r}_{2} \left(\bar{r}_{2} \cdot \hat{J}_{1}\right)}{I_{2}}\right) \cdot \hat{n}}$$
$$= \frac{-(1+\varepsilon) \cdot \bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n}}{\left(\hat{J}_{1} \cdot \hat{n}\right) \left(\frac{1}{\bar{m}} + \frac{|\bar{r}_{1}|^{2}}{I_{1}} + \frac{|\bar{r}_{2}|^{2}}{I_{2}}\right) - \left(\frac{\left(\bar{r}_{1} \cdot \hat{n}\right) \cdot \left(\bar{r}_{1} \cdot \hat{J}_{1}\right)}{I_{1}} + \frac{\left(\bar{r}_{2} \cdot \hat{n}\right) \cdot \left(\bar{r}_{2} \cdot \hat{J}_{1}\right)}{I_{2}}\right)}$$
$$= \frac{-(1+\varepsilon) \cdot \left(\bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n}\right) \cdot \sqrt{1+\mu^{2}}}{\left(\frac{1}{\bar{m}} + \frac{|\bar{r}_{1}|^{2}}{I_{1}} + \frac{|\bar{r}_{2}|^{2}}{I_{2}}\right) - \sqrt{1+\mu^{2}} \left(\frac{\left(\bar{r}_{1} \cdot \hat{n}\right) \cdot \left(\bar{r}_{1} \cdot \hat{J}_{1}\right)}{I_{1}} + \frac{\left(\bar{r}_{2} \cdot \hat{n}\right) \cdot \left(\bar{r}_{2} \cdot \hat{J}_{1}\right)}{I_{2}}\right)} \tag{46}$$
We can express the moment-arms in vector form by:
$$\bar{r}_{1} = r_{1n} \hat{n} + r_{1t} \hat{t} \tag{47}$$
and
$$\bar{r}_{2} = r_{2n} \hat{n} + r_{2t} \hat{t} \tag{48}$$
Therefore the inner products in (46) are:
$$\bar{r}_{1} \cdot \hat{J}_{1} = \frac{r_{1n} + \mu r_{1t}}{\sqrt{1+\mu^{2}}} \tag{49}$$
and
$$\bar{r}_{2} \cdot \hat{J}_{1} = \frac{r_{2n} + \mu r_{2t}}{\sqrt{1+\mu^{2}}} \tag{50}$$
We can now simplify (46) by:
$$|\bar{J}_{1}| = \frac{-(1+\varepsilon) \cdot (\bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n}) \cdot \sqrt{1+\mu^{2}}}{\left(\frac{1}{\bar{m}} + \frac{|\bar{r}_{1}|^{2}}{I_{1}} + \frac{|\bar{r}_{2}|^{2}}{I_{2}}\right) - \left(\frac{r_{1n}^{2}}{I_{1}} + \frac{r_{2n}^{2}}{I_{2}}\right) - \mu \left(\frac{r_{1n} r_{1t}}{I_{1}} + \frac{r_{2n} r_{2t}}{I_{2}}\right)} \tag{51}$$
where we have our primary results for object 1:
$$\bar{J}_{1} \cdot \hat{n} = \frac{-(1+\varepsilon) \cdot (\bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n})}{\left(\frac{1}{\bar{m}} + \frac{r_{1t}^{2}}{I_{1}} + \frac{r_{2t}^{2}}{I_{2}}\right) - \mu \left(\frac{r_{1n} r_{1t}}{I_{1}} + \frac{r_{2n} r_{2t}}{I_{2}}\right)} \tag{52}$$
$$\bar{J}_{1} \cdot \hat{t} = \frac{-\mu \cdot (1+\varepsilon) \cdot (\bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n})}{\left(\frac{1}{\bar{m}} + \frac{r_{1t}^{2}}{I_{1}} + \frac{r_{2t}^{2}}{I_{2}}\right) - \mu \left(\frac{r_{1n} r_{1t}}{I_{1}} + \frac{r_{2n} r_{2t}}{I_{2}}\right)} \tag{53}$$
For object 2 we have:
$$\bar{J}_{2} \cdot \hat{n} = \frac{(1+\varepsilon) \cdot (\bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n})}{\left(\frac{1}{\bar{m}} + \frac{r_{1t}^{2}}{I_{1}} + \frac{r_{2t}^{2}}{I_{2}}\right) - \mu \left(\frac{r_{1n} r_{1t}}{I_{1}} + \frac{r_{2n} r_{2t}}{I_{2}}\right)} \tag{54}$$
$$\bar{J}_{2} \cdot \hat{t} = \frac{\mu \cdot (1+\varepsilon) \cdot (\bar{v}_{\text{Rel,i}}^{P} \cdot \hat{n})}{\left(\frac{1}{\bar{m}} + \frac{r_{1t}^{2}}{I_{1}} + \frac{r_{2t}^{2}}{I_{2}}\right) - \mu \left(\frac{r_{1n} r_{1t}}{I_{1}} + \frac{r_{2n} r_{2t}}{I_{2}}\right)} \tag{55}$$
Thus, once the normal axis, \(\mu\), moment-arms, and the closing-velocity vector are known, everything else can be determined. Virtual CRASH uses the vehicle geometry to determine the geometrical parameters such as the normal axis direction, whereas the user supplies \(\varepsilon\) and the maximum preferred \(\mu\).
Common Velocity Along \(\hat{t}\)
During the compression phase, care is taken to ensure that the relative velocity along the tangent axis is not effectively accelerated by dissipative frictional forces beyond the point where the surfaces of contact reach common velocity along \(\hat{t}\); therefore, in Virtual CRASH the maximum allowed values of \(\bar{J}_{1} \cdot \hat{t}\) and \(\bar{J}_{2} \cdot \hat{t}\) are found to protect against this possibility. Here we derive the procedure.
Let us rewrite (37) as:
$$ \bar{v}_{\text{Rel,f}}^{P} - \bar{v}_{\text{Rel,i}}^{P} = \bar{J}_{1} \left(\frac{1}{\bar{m}} + \frac{|\bar{r}_{1}|^{2}}{I_{1}} + \frac{|\bar{r}_{2}|^{2}}{I_{2}}\right) - \left(\frac{\bar{r}_{1} \left(\bar{r}_{1} \cdot \bar{J}_{1}\right)}{I_{1}} + \frac{\bar{r}_{2} \left(\bar{r}_{2} \cdot \bar{J}_{1}\right)}{I_{2}}\right) \tag{56} $$
Projecting this on \(\hat{t}\) gives:
$$\left(\bar{v}_{\text{Rel,f}}^{P} - \bar{v}_{\text{Rel,i}}^{P}\right) \cdot \hat{t} = \left(\bar{J}_{1} \cdot \hat{t}\right) \left(\frac{1}{\bar{m}} + \frac{|\bar{r}_{1}|^{2}}{I_{1}} + \frac{|\bar{r}_{2}|^{2}}{I_{2}}\right) - \left(\frac{\left(\bar{r}_{1} \cdot \hat{t} \right) \left(\bar{r}_{1} \cdot \bar{J}_{1}\right)}{I_{1}} + \frac{\left(\bar{r}_{2} \cdot \hat{t} \right) \left(\bar{r}_{2} \cdot \bar{J}_{1}\right)}{I_{2}}\right)$$
$$= \frac{\mu \cdot |\bar{J}_{1}|}{\sqrt{1+\mu^{2}}} \left(\frac{1}{\bar{m}} + \frac{r_{1n}^{2}}{I_{1}} + \frac{r_{2n}^{2}}{I_{2}}\right) - \frac{|\bar{J}_{1}|}{\sqrt{1+\mu^{2}}} \left[\frac{r_{1t} r_{1n}}{I_{1}} + \frac{r_{2t} r_{2n}}{I_{2}}\right] \tag{57}$$
We can now solve for the value of \(\tilde{\mu}\) that will satisfy the common tangential velocity condition at point P:
$$ \bar{v}_{Rel,f}^P \cdot \hat{t} = 0 \tag{58}$$
This is given by:
$$- \bar{v}_{Rel,i}^P \cdot \hat{t} =\left(\frac{\tilde{\mu} |\bar{J}_1|}{\sqrt{1+\tilde{\mu}^2}}\right) \left(\frac{1}{\bar{m}} +\frac{r_{1n}^2}{I_1} +\frac{r_{2n}^2}{I_2}\right) - \frac{|\bar{J}_1|}{\sqrt{1+\tilde{\mu}^2}} \left[\frac{r_{1t} r_{1n}}{I_1} +\frac{r_{2t} r_{2n}}{I_2}\right] \tag{59}$$
Simplifying this gives:
$$\frac{\left(- \bar{v}_{Rel,i}^P \cdot \hat{t}\right) \sqrt{1+\tilde{\mu}^2}}{|\bar{J}_1|} = \tilde{\mu}\left(\frac{1}{\bar{m}} + \frac{r_{1n}^2}{I_1} + \frac{r_{2n}^2}{I_2}\right) - \left[ \frac{r_{1t} r_{1n}}{I_1} + \frac{r_{2t} r_{2n}}{I_2} \right] $$
$$= \frac{\left(\frac{1}{\bar{m}} + \frac{r_{1t}^2}{I_1} + \frac{r_{2t}^2}{I_2} \right) - \mu\left(\frac{r_{1n} r_{1t}}{I_1} + \frac{r_{2n} r_{2t}}{I_2}\right)}{-\left(\frac{(1+\varepsilon) \cdot (\bar{v}_{Rel,i}^P \cdot \hat{n})}{-\bar{v}_{Rel,i}^P \cdot \hat{t}}\right)} $$
$$= \tilde{\mu}\left(\frac{1}{\bar{m}} + \frac{r_{1n}^2}{I_1} + \frac{r_{2n}^2}{I_2}\right) - \left[ \frac{r_{1t} r_{1n}}{I_1} + \frac{r_{2t} r_{2n}}{I_2} \right] \tag{60}$$
which gives:
$$\left(\frac{1}{\bar{m}} +\frac{r_{1t}^2}{I_1} +\frac{r_{2t}^2}{I_2}\right)/\left(\frac{(1+\varepsilon) \cdot (\bar{v}_{Rel,i}^P \cdot \hat{n})}{\bar{v}_{Rel,i}^P \cdot \hat{t}}\right) + \left[\frac{r_{1n} r_{1t}}{I_1} +\frac{r_{2n} r_{2t}}{I_2}\right] = $$
$$\tilde{\mu}\left(\frac{1}{\bar{m}} +\frac{r_{1n}^2}{I_1} +\frac{r_{2n}^2}{I_2} + \left(\frac{r_{1n} r_{1t}}{I_1} +\frac{r_{2n} r_{2t}}{I_2}\right)/\left(\frac{(1+\varepsilon) \cdot (\bar{v}_{Rel,i}^P \cdot \hat{n})}{\bar{v}_{Rel,i}^P \cdot \hat{t}}\right)\right) \tag{61}$$
Finally, setting \(\varepsilon=0\) to evaluate our expression at the end of compression (just before restitution), and solving for \(\tilde{\mu}\) and simplifying, we have:
$$\tilde{\mu} = \frac{\frac{r_{1n} r_{1t}}{I_1} + \frac{r_{2n} r_{2t}}{I_2} + \left(\frac{\bar{v}_{Rel,i}^P \cdot \hat{t}}{\bar{v}_{Rel,i}^P \cdot \hat{n}}\right) \left(\frac{1}{\bar{m}} + \frac{r_{1t}^2}{I_1} + \frac{r_{2t}^2}{I_2}\right)}{\frac{1}{\bar{m}} + \frac{r_{1n}^2}{I_1} + \frac{r_{2n}^2}{I_2} + \left(\frac{\bar{v}_{Rel,i}^P \cdot \hat{t}}{\bar{v}_{Rel,i}^P \cdot \hat{n}}\right) \left(\frac{r_{1n} r_{1t}}{I_1} + \frac{r_{2n} r_{2t}}{I_2}\right)} \tag{62}$$
In general, the sign of \(\mu\) is set such that
$$sign(\mu) = sign\left(\frac{\bar{v}_{Rel,i}^P \cdot \hat{t}}{\bar{v}_{Rel,i}^P \cdot \hat{n}}\right) \tag{63} $$
This ensures that the magnitude of the tangent relative-velocity can only be reduced in value by friction, with maximum changes resulting in common velocity along the tangent axis direction at the end of compression.
Therefore, during compression the impulse along the tangent axis direction is bounded by the condition:
$$|\bar{J} \cdot \hat{t}| \leq |\tilde{\mu}(\bar{J} \cdot \hat{n})| \tag{64}$$
In Virtual CRASH, the user defines the preferred coefficient-of-friction used by the collision model. Each collision is checked against this condition to ensure the tangent impulse is properly bounded. For a given value of \(\mu\) specified by the user \(\left(\mu_{\text{user}}\right)\), Virtual CRASH will display "friction cones" about the collision point in order to illustrate the volume in which the impulse vectors, \(\bar{J}\), must be contained for each object. The size of the cone is determined by the user’s choice of \(\mu\); however, the effective cone may actually be smaller in order to satisfy equation (64).
With \(\varepsilon=0\) (as defined by (37) above), the post-impact common velocity condition is satisfied along the \(\hat{n}\) axis \(\left(\bar{v}_{\text{Rel,f}}^{P} \cdot \hat{n} = 0\right)\). Along the \(\hat{t}\) axis, \(\left(\bar{v}_{\text{Rel,f}}^{P} \cdot \hat{t}\right)\) will depend on \(\mu = \min(\tilde{\mu}, \mu_{\text{user}})\). With this, both components of \(\bar{J}\), \(\left(\bar{J} \cdot \hat{n}\right)\) and \(\left(\bar{J} \cdot \hat{t}\right)\), are determined for the compression phase of contact. The total impulse, accounting for both the compression phase and restitution phase, is then given by the product \((1+\varepsilon_{\text{user}}) \cdot \bar{J}\), where \(\varepsilon_{\text{user}}\) is the user-input restitution into the collision model.
Note, the collision model illustrated above is generalized in three-dimensions in Virtual CRASH such that the point of application of impulse vectors (impulse centroid) can be anywhere within the volume of the interacting objects, with impulse vectors pointing out of the interacting objects’ local x-y planes. In the general case, the impulse on object 1 is given by:
$$\bar{J}_1 = -(1+\varepsilon)\left[\left(\frac{1}{\bar{m}}\right)\mathbf{1} - \left(\tilde{\mathbf{r}}_1 \mathbf{I}_1^{-1} \tilde{\mathbf{r}}_1 + \tilde{\mathbf{r}}_2 \mathbf{I}_2^{-1} \tilde{\mathbf{r}}_2\right)\right]^{-1} \cdot \bar{v}_{\text{Rel,i}}^{P}$$
where \(\mathbf{\tilde{r}_1}\) and \(\mathbf{\tilde{r}_2}\) are the vectors extending from the centers-of-gravity of object 1 and 2, respectively, to the point-of-contact, \(P\). These vectors are represented in skew-symmetric matrix representation. \(\mathbf{I}_1\) and \(\mathbf{I}_2\) are moment-of-inertia tensors for objects 1 and 2, respectively, \(\varepsilon\) is the coefficient-of-restitution, \(\bar{m}\) is the system’s reduced mass, and \(\mathbf{1}\) is the 3x3 identity matrix. The impulse on object 2 is then given by \(\bar{J}_2 = -\bar{J}_1\) as required by Newton’s 3rd Law. The impulses are converted to equivalent contact forces by \(\bar{F} = \bar{J} / \delta t\), where \(\delta t\) is one integration time-step. For each time-step, and for each vehicle, contact forces are vector-summed with all other forces acting on the given vehicle.
Virtual CRASH numerically solves the Newton-Euler equations to evolve object linear and angular velocities, as well as positions and orientations, forward with time.
References
[1] Slibar A., “ Die mechanischen GmndsXtze des StoBvorganges freier und gefuhrter Korper und ihre Anwendung auf den StoBvorgang van Fabrzeugen.” Archiv for Unfallforschung, 2. Jg., H. 1, 1966, 31ff.
[5] See for example, the Planar Impact Mechanics Analysis software module of VCRWare by Brach Engineering (brachengineering.com), and the associated SAE Text: “Vehicle Accident Analysis and Reconstruction Methods”, R. Brach and R. Brach, SAE International, Warrendale, Pennsylvania, 2005.