# Convert gram / barrel to pound / cubic yard

Learn how to convert 1 gram / barrel to pound / cubic yard step by step.

## Calculation Breakdown

Set up the equation
$$1.0\left(\dfrac{gram}{barrel}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{pound}{cubic \text{ } yard}\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)$$
$$\text{Left side: 1.0 } \left(\dfrac{gram}{barrel}\right) = {\color{rgb(89,182,91)} \dfrac{10^{-2}}{1.6365924}\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(89,182,91)} \dfrac{10^{-2}}{1.6365924}\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}$$
$$\text{Right side: 1.0 } \left(\dfrac{pound}{cubic \text{ } yard}\right) = {\color{rgb(125,164,120)} \dfrac{5.0}{7.64554857984}\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(125,164,120)} \dfrac{5.0}{7.64554857984}\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(\dfrac{gram}{barrel}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{pound}{cubic \text{ } yard}\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{10^{-2}}{1.6365924}} \times {\color{rgb(89,182,91)} \left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{5.0}{7.64554857984}}} \times {\color{rgb(125,164,120)} \left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} \dfrac{10^{-2}}{1.6365924}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{5.0}{7.64554857984}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{10^{-2}}{1.6365924}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{5.0}{7.64554857984}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}}$$
$$\text{Conversion Equation}$$
$$\dfrac{10^{-2}}{1.6365924} = {\color{rgb(20,165,174)} x} \times \dfrac{5.0}{7.64554857984}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{5.0}{7.64554857984} = \dfrac{10^{-2}}{1.6365924}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{7.64554857984}{5.0}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{5.0}{7.64554857984} \times \dfrac{7.64554857984}{5.0} = \dfrac{10^{-2}}{1.6365924} \times \dfrac{7.64554857984}{5.0}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{5.0}} \times {\color{rgb(99,194,222)} \cancel{7.64554857984}}}{{\color{rgb(99,194,222)} \cancel{7.64554857984}} \times {\color{rgb(255,204,153)} \cancel{5.0}}} = \dfrac{10^{-2} \times 7.64554857984}{1.6365924 \times 5.0}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{10^{-2} \times 7.64554857984}{1.6365924 \times 5.0}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx0.0093432532\approx9.3433 \times 10^{-3}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{gram}{barrel}\right)\approx{\color{rgb(20,165,174)} 9.3433 \times 10^{-3}}\left(\dfrac{pound}{cubic \text{ } yard}\right)$$