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Appendix A5 317

A5 Conversion of Units to SI

The following relationships can be used to convert customary physical units into SI units. The conversion factors here are rounded to seven significant digits.

l(m)=39.37008l(in)(A5/1)l(m) = 39.370 08 l(in) (A5/1)
l(in)=2.54×102l(m)(A5/2)l(in) = 2.54 \times 10^{-2}l(m) (A5/2)
l(m)=3.28084l(ft)(A5/3)l(m) = 3.280 84 l(ft) (A5/3)
l(ft)=0.3048l(m)(A5/4)l(ft) = 0.3048 l(m) (A5/4)
l(m)=6.213712×104l(mi)(A5/5)l(m) = 6.213 712 \times 10^{-4}l(mi) (A5/5)
l(mi)=1.609344×103l(m)(A5/6)l(mi) = 1.609 344 \times 10^{3}l(m) (A5/6)
C(ms1)=0.44704C(mph)(A5/7)C(m s^{-1}) = 0.447 04 C(mph) (A5/7)
C(km/h)=1.609344C(mph)(A5/8)C(km/h) = 1.609 344 C(mph) (A5/8)
C(mph)=0.621371C(km/h)(A5/9)C(mph) = 0.621 371 C(km/h) (A5/9)
C(mph)=2.236936C(ms1)(A5/10)C(mph) = 2.236 936 C(m s^{-1}) (A5/10)
V(USgal)=2.641721×102V(m3)(A5/11)V(US gal) = 2.641 721 \times 10^{2}V(m^{3}) (A5/11)
V(m3)=3.785412×103V(USgal)(A5/12)V(m^{3}) = 3.785 412 \times 10^{-3}V(US gal) (A5/12)
V(impgal)=2.199692×102V(m3)(A5/13)V(imp gal) = 2.199 692 \times 10^{2}V(m^{3}) (A5/13)
V(m3)=4.54609×103V(impgal)(A5/14)V(m^{3}) = 4.546 09 \times 10^{-3}V(imp gal) (A5/14)
m(t)=103m(kg)(A5/15)m(t) = 10^{-3}m(kg) (A5/15)
m(kg)=103m(t)(A5/16)m(kg) = 10^{3}m(t) (A5/16)
m(lb)=2.204623m(kg)(A5/17)m(lb) = 2.204 623 m(kg) (A5/17)
m(kg)=0.45359237m(lb)(A5/18)m(kg) = 0.453 592 37 m(lb) (A5/18)

318 Appendix A5

F(lbf)=0.2248089F(N)(A5/19)F(lbf) = 0.224 808 9 F(N) (A5/19)
F(N)=4.448222F(lbf)(A5/20)F(N) = 4.448 222 F(lbf) (A5/20)
p(inHg)=2.067099×106p(Pa)(A5/21)p(inHg) = 2.067 099 \times 10^{-6}p(Pa) (A5/21)
p(Pa)=3.386389×103p(inHg)(A5/22)p(Pa) = 3.386 389 \times 10^{3}p(inHg) (A5/22)
p(bar)=105p(Pa)(A5/23)p(bar) = 10^{-5}p(Pa) (A5/23)
p(Pa)=105p(bar)(A5/24)p(Pa) = 10^{5}p(bar) (A5/24)
p(psi)=1.450377×104p(Pa)(A5/25)p(psi) = 1.450 377 \times 10^{-4}p(Pa) (A5/25)
p(psi)=14.50377p(bar)(A5/26)p(psi) = 14.503 77 p(bar) (A5/26)
p(Pa)=6.894757×103p(psi)(A5/27)p(Pa) = 6.894 757 \times 10^{3}p(psi) (A5/27)
p(bar)=6.894757×102p(psi)(A5/28)p(bar) = 6.894 757 \times 10^{-2}p(psi) (A5/28)
p(kgf/cm2)=1.019716×105p(Pa)(A5/29)p(kg_{f}/cm^{2}) = 1.019 716 \times 10^{-5}p(Pa) (A5/29)
p(kgf/cm2)=1.019716p(bar)(A5/30)p(kg_{f}/cm^{2}) = 1.019 716 p(bar) (A5/30)
p(Pa)=9.80665×104p(kgf/cm2)(A5/31)p(Pa) = 9.806 65 \times 10^{4}p(kg_{f}/cm^{2}) (A5/31)
p(bar)=0.980665p(kgf/cm2)(A5/32)p(bar) = 0.980 665 p(kg_{f}/cm^{2}) (A5/32)
T(F)=1.8×[T(K)273.15]+32(A5/33)T(^{\circ} F) = 1.8 \times [T(K) - 273.15] + 32 (A5/33)
T(K)=T(F)321.8+273.15(A5/34)T(K) = \frac{T(^{\circ} F) - 32}{1.8} + 273.15 (\mathrm{A}5/34)
W(hpimp)=1.341022×103W˙(W)(A5/35)W(hp imp) = 1.341 022 \times 10^{-3}\dot{W}(W) (A5/35)
W(W)=7.456999×102W˙(hpimp)(A5/36)W(W) = 7.456 999 \times 10^{2}\dot{W}(hp imp) (A5/36)
W(hp)=1.359622×103W˙(W)(A5/37)W(hp) = 1.359 622 \times 10^{-3}\dot{W}(W) (A5/37)
W(W)=7.354988×102W˙(hp)(A5/38)W(W) = 7.354 988 \times 10^{2}\dot{W}(hp) (A5/38)