09-28-2021, 08:53 AM
Sure that here, in that HP calculator forum, is not the normal place to place my question.
But that interrogation occurred now when developing my gravity program for the HP49-HP50.
And as I know that the level of some HP user is quite high, counting a lot of maths people or ingenieurs, I dare and submit my problem, hoping for your indulgence... and help.
In the Wikipedia "Theoretical Gravity" article,
choose the Chinese language.
There is to be found a formula j.e
(indeed j.p) for the gravity g at the pole.
j.pole =
'GM/(a*a)*(1+m+3/7*e'²*m)' equation 1
In WGS84
a=exactly 6 378 137
m='w*w*a*a*b/GM'
with w: = exactly .00007292115
b=a-a*f and f=exactly 298.257223563
so that b=about 6356752. 3142451794975639665996336551567981713110854973388485716512832085220868351009394057094507132374770633
And GM=exactly 3.986004418E14
so that m=about 0.0034497865068408453070274965050786700048416482688355349290136425040253035757396870475700989923885167801669828495872298265101288596701656992493579318456239603696294749064174268559478550984385788003910837612
e'²='(a*a-b*b)/(b*b)'
so that e'²=about 0.0067394967422764349547821589567593766561220658588440099819050282583227973208914914189860809667349634
So that (1+m+3/7*e'²*m)=
1.003459750717522732340985895556610125671094437673903614541154381375001105833623894718788389499178806618567527413384953945229190157855623901712950238982395888274759708044521808294166478667460951510991749895895279214598597362085502307407034549091368688099259583762966067132573436092217775000400440672864145748571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571
And finally g. Pole (without the dot/comma) should be exactly equal to, according to equation 1:
'3986004418*1003459750717522732340985895556610125671094437673903614541154381375001105833623894718788389499178806618567527413384953945229190157855623901712950238982395888274759708044521808294166478667460951510991749895895279214598597362085502307407034549091368688099259583762966067132573436092217775000400440672864145748571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571/(6378137*6378137)'
Or j. Pole, placing back the dot/comma at the right place=
9.8321851044044354166235845241110015889520125854466337942802305915361947992868863041785670259867943125529117241157228655849590168522129663217533159804636091817559953695048056826641872205769685950655291862172166302389558892330844759850013860675472423221623311315941998400543951488023702988151342889690235373769066691915076944701814198886293763195146108008175411982135184646619151544169766231068819297264703012453. (Full digit calculated result)
But all the formulae given for the gravity at the pole write that it is exactly? =
9.83218 49378 (General given 11 digits result)
My questions
Does a term, or several, miss in the equation 1?
Or, on the the contrary, Full digit calculated result is correct and General given 11 digits result was I then incorrectly rounded (with the 7th digit already false)?
As a curious layman, I would appreciate your precious insights.
Thanks for your help.
Regards,
Gil Campart
But that interrogation occurred now when developing my gravity program for the HP49-HP50.
And as I know that the level of some HP user is quite high, counting a lot of maths people or ingenieurs, I dare and submit my problem, hoping for your indulgence... and help.
In the Wikipedia "Theoretical Gravity" article,
choose the Chinese language.
There is to be found a formula j.e
(indeed j.p) for the gravity g at the pole.
j.pole =
'GM/(a*a)*(1+m+3/7*e'²*m)' equation 1
In WGS84
a=exactly 6 378 137
m='w*w*a*a*b/GM'
with w: = exactly .00007292115
b=a-a*f and f=exactly 298.257223563
so that b=about 6356752. 3142451794975639665996336551567981713110854973388485716512832085220868351009394057094507132374770633
And GM=exactly 3.986004418E14
so that m=about 0.0034497865068408453070274965050786700048416482688355349290136425040253035757396870475700989923885167801669828495872298265101288596701656992493579318456239603696294749064174268559478550984385788003910837612
e'²='(a*a-b*b)/(b*b)'
so that e'²=about 0.0067394967422764349547821589567593766561220658588440099819050282583227973208914914189860809667349634
So that (1+m+3/7*e'²*m)=
1.003459750717522732340985895556610125671094437673903614541154381375001105833623894718788389499178806618567527413384953945229190157855623901712950238982395888274759708044521808294166478667460951510991749895895279214598597362085502307407034549091368688099259583762966067132573436092217775000400440672864145748571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571
And finally g. Pole (without the dot/comma) should be exactly equal to, according to equation 1:
'3986004418*1003459750717522732340985895556610125671094437673903614541154381375001105833623894718788389499178806618567527413384953945229190157855623901712950238982395888274759708044521808294166478667460951510991749895895279214598597362085502307407034549091368688099259583762966067132573436092217775000400440672864145748571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571428571/(6378137*6378137)'
Or j. Pole, placing back the dot/comma at the right place=
9.8321851044044354166235845241110015889520125854466337942802305915361947992868863041785670259867943125529117241157228655849590168522129663217533159804636091817559953695048056826641872205769685950655291862172166302389558892330844759850013860675472423221623311315941998400543951488023702988151342889690235373769066691915076944701814198886293763195146108008175411982135184646619151544169766231068819297264703012453. (Full digit calculated result)
But all the formulae given for the gravity at the pole write that it is exactly? =
9.83218 49378 (General given 11 digits result)
My questions
Does a term, or several, miss in the equation 1?
Or, on the the contrary, Full digit calculated result is correct and General given 11 digits result was I then incorrectly rounded (with the 7th digit already false)?
As a curious layman, I would appreciate your precious insights.
Thanks for your help.
Regards,
Gil Campart