# Convert degree / square hour to cycle / square hour

Learn how to convert 1 degree / square hour to cycle / square hour step by step.

## Calculation Breakdown

Set up the equation
$$1.0\left(\dfrac{degree}{square \text{ } hour}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{cycle}{square \text{ } hour}\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{radian}{square \text{ } second}\right)$$
$$\text{Left side: 1.0 } \left(\dfrac{degree}{square \text{ } hour}\right) = {\color{rgb(89,182,91)} \dfrac{π}{2.3328 \times 10^{9}}\left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(89,182,91)} \dfrac{π}{2.3328 \times 10^{9}}\left(\dfrac{rad}{s^{2}}\right)}$$
$$\text{Right side: 1.0 } \left(\dfrac{cycle}{square \text{ } hour}\right) = {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}\left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}\left(\dfrac{rad}{s^{2}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(\dfrac{degree}{square \text{ } hour}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{cycle}{square \text{ } hour}\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{2.3328 \times 10^{9}}} \times {\color{rgb(89,182,91)} \left(\dfrac{radian}{square \text{ } second}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}}} \times {\color{rgb(125,164,120)} \left(\dfrac{radian}{square \text{ } second}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} \dfrac{π}{2.3328 \times 10^{9}}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{rad}{s^{2}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{rad}{s^{2}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{2.3328 \times 10^{9}}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{rad}{s^{2}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{π}{6.48 \times 10^{6}}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{rad}{s^{2}}\right)}}$$
$$\text{Conversion Equation}$$
$$\dfrac{π}{2.3328 \times 10^{9}} = {\color{rgb(20,165,174)} x} \times \dfrac{π}{6.48 \times 10^{6}}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$$\dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{2.3328 \times {\color{rgb(99,194,222)} \cancelto{10^{3}}{10^{9}}}} = {\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{6.48 \times {\color{rgb(99,194,222)} \cancel{10^{6}}}}$$
$$\text{Simplify}$$
$$\dfrac{1.0}{2.3328 \times 10^{3}} = {\color{rgb(20,165,174)} x} \times \dfrac{1.0}{6.48}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{6.48} = \dfrac{1.0}{2.3328 \times 10^{3}}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{6.48}{1.0}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{6.48} \times \dfrac{6.48}{1.0} = \dfrac{1.0}{2.3328 \times 10^{3}} \times \dfrac{6.48}{1.0}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times {\color{rgb(99,194,222)} \cancel{6.48}}}{{\color{rgb(99,194,222)} \cancel{6.48}} \times {\color{rgb(255,204,153)} \cancel{1.0}}} = \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times 6.48}{2.3328 \times 10^{3} \times {\color{rgb(255,204,153)} \cancel{1.0}}}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{6.48}{2.3328 \times 10^{3}}$$
Rewrite equation
$$\dfrac{1.0}{10^{3}}\text{ can be rewritten to }10^{-3}$$
$$\text{Rewrite}$$
$${\color{rgb(20,165,174)} x} = \dfrac{10^{-3} \times 6.48}{2.3328}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx0.0027777778\approx2.7778 \times 10^{-3}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{degree}{square \text{ } hour}\right)\approx{\color{rgb(20,165,174)} 2.7778 \times 10^{-3}}\left(\dfrac{cycle}{square \text{ } hour}\right)$$