# Analysis of Underlapped Symmetrically Ported Valve-Controlled Asymmetric Cylinder Drive
## 3 Materials and Methods
### Derivation of Analytical Solution for the Value of \(X\) Factor  
Squaring both sides of Equation (25) leads to:  
\[ \begin{split}&\frac{A_{V1}}{\gamma^{2}}\big{(}1+\gamma^{3}-X\big{)}+ \frac{A_{V2}}{\gamma^{2}}X-\frac{2A_{V1}A_{V2}}{\gamma^{2}}\big{(}X+\gamma^{3} X-X^{2}\big{)}^{\frac{1}{2}}\\ &=A_{V1}{}^{2}X\gamma+A_{V2}{}^{2}\big{(}1+\gamma^{3}-X\gamma \big{)}-2A_{V1}A_{V2}\big{(}X\gamma+X\gamma^{4}-X^{2}\gamma^{2}\big{)}^{\frac {1}{2}}\end{split}  \tag{28}\]  
Rearranged:  
\[ \begin{split}&\Big{[}\frac{A_{V1}}{2A_{V2}}\big{(}1+\gamma^{3} \big{)}-\frac{A_{V2}}{2A_{V1}}\big{(}\gamma^{2}+\gamma^{5}\big{)}\Big{]}- \Big{[}\frac{A_{V1}}{2A_{V2}}\big{(}1+\gamma^{3}\big{)}-\frac{A_{V2}}{2A_{V1}} \big{(}1+\gamma^{3}\big{)}\Big{]}X\\ &=\big{[}\big{(}1+\gamma^{3}\big{)}X-X^{2}\big{]}^{\frac{1}{2}}- \gamma^{2}\big{[}\big{(}\gamma+\gamma^{4}\big{)}X-\gamma^{2}X^{2}\big{]}^{ \frac{1}{2}}.\end{split}  \tag{29}\]  
Let \(m\) be given as follows:  
\[m=\frac{A_{V1}}{2A_{V2}}\Big{(}1+\gamma^{3}\Big{)}-\frac{A_{V2}}{2A_{V1}} \Big{(}\gamma^{2}+\gamma^{5}\Big{)}=\Big{(}1+\gamma^{3}\Big{)}\bigg{(}\frac{ A_{V1}}{2A_{V2}}-\frac{A_{V2}}{2A_{V1}}\bigg{)} \tag{30}\]  
Let \(n\) be given as follows:  
\[n=\frac{A_{V1}}{2A_{V2}}\Big{(}1+\gamma^{3}\Big{)}-\frac{A_{V2}}{2A_{V1}} \Big{(}1+\gamma^{3}\Big{)}=\Big{(}1+\gamma^{3}\Big{)}\bigg{(}\frac{A_{V1}}{2A _{V2}}-\frac{A_{V2}}{2A_{V1}}\bigg{)} \tag{31}\]  
Substituting \(m\) and \(n\) into Equation (31) reveals the following:  
\[m-nX=\Big{[}\Big{(}1+\gamma^{3}\Big{)}X-X^{2}\Big{]}^{\frac{1}{2}}-\gamma^{2 }\Big{[}\Big{(}\gamma+\gamma^{4}\Big{)}X-\gamma^{2}X^{2}\Big{]}^{\frac{1}{2}} \tag{32}\]  
Figure 9: Chamber pressures when the valve is in underlap state. Squaring both sides of Equation (32) leads to:  
\[ \begin{array}{ll}m^{2}+n^{2}X^{2}-2 mnX&=\big{(}1+\gamma^{3}\big{)}X-X^{2}+\gamma^ {5}\big{(}1+\gamma^{3}\big{)}X-\gamma^{6}X^{2}\\ &-2\gamma^{2}\big{[}\gamma\big{(}1+\gamma^{3}\big{)}^{2}X^{2}-\gamma\big{(}1+ \gamma^{3}\big{)}X^{3}-\gamma^{2}\big{(}1+\gamma^{3}\big{)}X^{3}+\gamma^{2}X^{4 }\big{]}^{\frac{1}{2}}\end{array}  \tag{33}\]  
Rearranging Equation (33) reveals the following:  
\[ \begin{array}{ll}\big{(}n^{2}+1+\gamma^{6}\big{)}X^{2}&-\big{[}2 mn+\big{(}1 +\gamma^{5}\big{)}\big{(}1+\gamma^{3}\big{)}\big{]}X+m^{2}\\ &=2\gamma^{2}\Big{[}\gamma\big{(}1+\gamma^{3}\big{)}^{2}X^{2}-\gamma\big{(}1+ \gamma^{3}\big{)}X^{3}-\gamma^{2}\big{(}1+\gamma^{3}\big{)}X^{3}+\gamma^{2}X^{ 4}\Big{]}^{\frac{1}{2}}\end{array}  \tag{34}\]  
Let  
\[ \begin{array}{c}p=n^{2}+1+\gamma^{6}\\ q=2 mn+\big{(}1+\gamma^{5}\big{)}\big{(}1+\gamma^{3}\big{)}\\ h=m^{2}\end{array}  \tag{35}\]  
Equation (34) becomes the following:  
\[pX^{2}-qX+h=2\gamma^{2}\bigg{[}\gamma\Big{(}1+\gamma^{3}\Big{)}^{2}X^{2}- \gamma\Big{(}1+\gamma^{3}\Big{)}X^{3}-\gamma^{2}\Big{(}1+\gamma^{3}\Big{)}X^{ 3}+\gamma^{2}X^{4}\bigg{]}^{\frac{1}{2}} \tag{36}\]  
Squaring both sides of Equation (36) reveals the following:  
\[ \begin{array}{ll}p^{2}X^{4}+q^{2}X^{2}+h^{2}&-pqX^{3}+phX^{2}-pqX^{3}-qhX+ phX^{2}-qhX\\ &=4\gamma^{6}X^{4}-4\gamma^{5}\big{(}1+\gamma\big{)}\big{(}1+\gamma^{3}\big{)} X^{3}+4\gamma^{5}\big{(}1+\gamma^{3}\big{)}^{2}X^{2}\end{array}  \tag{37}\]  
Rearranging the above equation reveals the following:  
\[ \begin{array}{ll}\big{(}p^{2}-4\gamma^{6}\big{)}X^{4}&+\big{[}4\gamma^{5} \big{(}1+\gamma\big{)}\big{(}1+\gamma^{3}\big{)}-2 pq\big{]}X^{3}+\Big{[}\big{(} 2 ph+q^{2}\big{)}-4\gamma^{5}\big{(}1+\gamma^{3}\big{)}^{2}\Big{]}X^{2}\\ &-2 qhX+h^{2}=0\end{array}  \tag{38}\]  
Let  
\[ \begin{array}{c}a_{4}=p^{2}-4\gamma^{6}\\ a_{3}=4\gamma^{5}\big{(}1+\gamma\big{)}\big{(}1+\gamma^{3}\big{)}-2 pq\\ a_{2}=\big{(}2 ph+q^{2}\big{)}-4\gamma^{5}\big{(}1+\gamma^{3}\big{)}^{2}\\ a_{1}=-2 qh\\ a_{0}=h^{2}\end{array}  \tag{39}\]  
Thus Equation (38) becomes the following:  
\[a_{4}X^{4}+a_{3}X^{3}+a_{2}X^{2}+a_{1}X+a_{0}=0 \tag{40}\]  
Let  
\[ \begin{array}{c}a=\frac{a_{1}}{a_{4}},b=\frac{a_{2}}{a_{4}}\\ c=\frac{a_{1}}{a_{4}},d=\frac{a_{0}}{a_{4}}\end{array}