The original book equation for speed walking aerobic fitness points.
PT=(Dx60/T)-1)xD-1 e.g. (2x60/30)-1)x2-1=5 points Where PT=points,
D=distance(2miles), & T=time(30 minutes). . This is the 1st calculator
equation : SPM=((MIN+SEC/60)/(BLK x .435)-13)x60. Where SPM=seconds per
mile (based on a 13 minute mile). BLK= .435 of a mile the distance around my neighborhood
block. The variables are: SPM. MIN. SEC. BLK . . Lets say I went 3 times around the
block in 17 minutes and 25.31 seconds and my total walking time was 22 minutes.
Therefore; MIN=17 SEC=25.31 BLK=3 SPM=??? ANSWER is 21 seconds per mile
because its based on 13 minute mile I know my speed per mile is 13:21 . Now my 2nd
equation in my HP 17BII has to do with my total walking time. I used an equation for distance
to combine two equations into one: D=(T/(SPM/60+13). Time divided by speed = distance
therefore 22/13:21 (21/60+13) is the same as (SPM/60+13) the answer is 1.65 miles is the distance.
and the points PT=4.76 This is the answer you should get in the 2nd equation below
Therefore; PT=(T/(SPM/60+13)x60)/T-1x(T/(SPM/60+13)-1) is the same as PT=(Dx60/T)-1)xD-1
The variables are PT.. T.. SPM. When working out equation in a step by step procedure it only works
out properly up to this point PT=(T/(SPM/60+13)x60)/T-1 If you put in the equation up to this point
using T=22 (my total walking time was 22 minutes) and SPM=21 then PT=?? answer 3.49 is the right
answer up to this point. when I put the rest of the equation i.e. x(T/SPM/60+13)-1) in the answer is 3.85
for PT but the correct answer for PT is 4.76 as you can see (T/SPM/60+13) is repeated twice in the equation
because it represents D for distance based on the original book equation The HP 17BII is not interpreting this
aspect properly. How do I represent this equation so that this aspect is taken into account? Thanks! .