Thrust to weight

1. Aug 8, 2004

wolram

Sorry if this question is on the basic side, but i need some way
of calculating the amount of thrust needed to propel a vehicle
a quarter of a mile in 10 seconds.
this vehicle is not built yet, but will be as light as possible, at a
rough guess 300lbs with driver.

2. Aug 8, 2004

abercrombiems02

lets begin with newtons second law
sum F= ma the only problem is m is changing. Lets make one assumption to make this problem a million times simpler. Lets model the mass of the rocket as
Mo - kt. where k is mass consumption rate of the propellant and Mo is the inert mass of the rocket in addition to the initial propellant mass of the rocket, Thus the mass of the rocket at any given time will be Mi + Mp - kt.(Mo = Mi + Mp)
so the acceleration is then (Thrust - Weight)/(Mi + Mp - kt) another great thing about your problem is that the distance is only a quareter mile, therefore another great assumption would be to assume that g is constant. so to find velocity lets integrate acceleration with respect to t. and we obtain v = T/k*(ln(Mo - kt) - ln(Mo)) + Vo, lets assume Vo is 0 if the rocket is resting on the ground. it would also follow that -kt = Mp expelled when the rocket has exhausted all its fuel, therefore v = F/k*ln((Mo + Mp)/Mo) however T/k = Go*Isp Go is the gravity (9.8) and the Isp is the effeciency of the fuel, therefore our thrust is equal to Go*Isp*k = T, this proves higher Isps give a better thrust. so lets integrate v to get or displacement x. if the initial velocity and displacement are 0 then the solution is
x(t) = T/k*[((mo +kt)*ln(mo+kt)-mo*ln(mo))/k+(vo - 1)*Mp/k]

from here simply choose x = 400m mo is the mass of the rocket, mp is the mass of the propellant, typically mp accounts for around 80-90% of the initial mass of the rocket. for smaller rockets this value is closer to 80%, choose a given k (this is the fuel consumption rate in kg/s, an your initial velocity vo is equal to zero. if at time t all the fuel is consumed, then the formula simplifies to

x(t burnout) = T/k*[((mo - mp)*ln(mo - mp)-mo*ln(mo))/k+(vo - 1)*Mp/k]

after this point you can simply use projectile motion equations to find the position of the rocket after the fuel has been expended if air resistance is neglected. one thing to note here when choose values, we said T/k = go*Isp well to give u an idea of performance limits, a VERY VERY good conventional rocket engine has an ISP of about 400 s (the unit in Isp is seconds) so your best T/k ratio would be roughly 4000. Another assumption made with this model is that the rocket when firing its engines immdeatly acquires its maximum thrust. In reality there is usually a delta function used and this adds another step but this is dependent on combustion rates, and the effeciency of the fuel pumps if we are using liquid fuel. so if I were you i would use this formula and then tweak around with one last quantity

400 = T/k*[((140 + 10k)*ln(140+10k)-140*ln(140))/k - PMF*140/k]

now lets apply the constraints T/k < 4000 and PMF
(the propellant mass fraction) must lie somwhere from 0.8 to 0.9 or so would be typical values for a conventional rocket. If we want to minimize the thrust required, we need to have a small value of k as well and a low PMF, this is why lower PMF's are common in smaller rockets, so for practical purposes i would choose 0.8

400 = T/k*[((140 + 10k)*ln(140+10k)-140*ln(140))/k - 0.8*140/k]

Now choose some T's and k's until you find a combination that seems approapriate keeping in mind (T/k) < 4000

3. Aug 8, 2004

abercrombiems02

if you want a better report u can check out the website my group used for our aerospace design www.yehudiworks.com[/URL], there are some rocket calculations in the final report that is downloadable off the website

Last edited by a moderator: Apr 21, 2017
4. Aug 8, 2004

wolram

abercrombiems02
You guys are so good, i can make almost anything from plans, but when
math is involved, i am blind, would you be interested in my project?
maybe you think i am pathetic, but i do want to make a last statement
and i think this is my last chance.

5. Aug 8, 2004

krab

This is a physics question. Why is it in the Math forum?