An online conversion calculator of yards, feet, inches, and meters is presented along with examples and problems of applications and their solutions.
Reference: NIST Reference Guidelines
Convert \( 12 \text{ yd } 2 \text{ ft } 7 \text{ in} \) to meters and round the answer to the nearest unit.
Using the conversion rates (\( 1 \text{ yd} = 91.44 \, \text{cm} \), \( 1 \text{ ft} = 30.48 \, \text{cm} \), \( 1 \text{ in} = 2.54 \, \text{cm} \)):
\[ 12 \text{ yd } 2 \text{ ft } 7 \text{ in} = 12 \times 91.44 + 2 \times 30.48 + 7 \times 2.54 = 1176.02 \, \text{cm} \]Convert centimeters to meters (\( 1 \, \text{cm} = 0.01 \, \text{m} \)):
\[ 1176.02 \, \text{cm} = 1176.02 \times 0.01 = 11.7602 \, \text{m} \]Rounding to the nearest unit:
\[ 12 \, \text{m} \]Convert \( 15.25 \, \text{meters} \) to yards, feet, and inches.
Convert meters to centimeters:
\[ 15.25 \, \text{m} = 15.25 \times 100 = 1525 \, \text{cm} \]Convert centimeters to yards (\( 1 \, \text{yd} = 91.44 \, \text{cm} \)):
\[ \dfrac{1525}{91.44} \approx 16.677603 \, \text{yd} = 16 \, \text{yd} + 0.677603 \, \text{yd} \]Convert the fractional yard part to feet (\( 1 \, \text{yd} = 3 \, \text{ft} \)):
\[ 0.677603 \times 3 = 2.032808 \, \text{ft} = 2 \, \text{ft} + 0.032808 \, \text{ft} \]Convert the fractional foot part to inches (\( 1 \, \text{ft} = 12 \, \text{in} \)):
\[ 0.032808 \times 12 \approx 0.3937 \, \text{in} \]Rounding to the nearest inch:
\[ 16 \text{ yd } 2 \text{ ft } 0 \text{ in} \]Enter the number of yd, ft, in, or the number of meters to perform conversions.
It costs \( \$12.5 \) per (linear) meter to fence a rectangular field of length \( L = 25 \text{ yd } 2 \text{ ft} \) and width \( W = 20 \text{ yd } 1 \text{ ft} \). What is the total cost of fencing the field along its perimeter?
Convert length and width into meters:
\[ L = 25 \times 91.44 + 2 \times 30.48 = 2286 + 60.96 = 2346.96 \, \text{cm} = 23.4696 \, \text{m} \] \[ W = 20 \times 91.44 + 1 \times 30.48 = 1828.8 + 30.48 = 1859.28 \, \text{cm} = 18.5928 \, \text{m} \]Calculate the perimeter \( P \):
\[ P = 2(L + W) = 2(23.4696 + 18.5928) = 84.1248 \, \text{m} \]Calculate the total cost:
\[ \text{Total Cost} = 84.1248 \, \text{m} \times \$12.5/\text{m} = \$1051.56 \]Convert the speed of \( 121.5 \, \text{yd/sec} \) into \( \text{m/sec} \) and round the answer to one decimal place.
Since \( 1 \, \text{yd} = 91.44 \, \text{cm} = 0.9144 \, \text{m} \):
\[ 121.5 \, \text{yd/sec} = 121.5 \times 0.9144 \, \text{m/sec} = 111.0996 \, \text{m/sec} \]Rounding to one decimal place:
\[ 111.1 \, \text{m/sec} \]A cubic water tank with 6 square faces has \( 1.6 \, \text{yd} \) per side (of each square). How long does it take to fill the tank with water at the rate of \( 0.5 \, \text{m}^3/\text{hour} \)?
Convert \( 1.6 \, \text{yd} \) into meters:
\[ 1.6 \, \text{yd} = 1.6 \times 0.9144 \, \text{m} = 1.46304 \, \text{m} \]Calculate the volume \( V \) of the cube in \( \text{m}^3 \):
\[ V = 1.46304 \times 1.46304 \times 1.46304 \approx 3.131617 \, \text{m}^3 \]Calculate the time \( T \) required to fill the tank at \( 0.5 \, \text{m}^3/\text{hr} \):
\[ T = \dfrac{3.131617 \, \text{m}^3}{0.5 \, \text{m}^3/\text{hr}} = 6.263233 \, \text{hours} \]Convert decimal hours into hours, minutes, and seconds:
\[ 6 \text{ hours} + 0.263233 \times 60 \text{ min} = 6 \text{ hours } 15.794 \text{ min} \] \[ 15 \text{ min} + 0.794 \times 60 \text{ sec} \approx 6 \text{ hours } 15 \text{ minutes } 48 \text{ seconds} \]