Calculate Thrust Needed for 300lb Vehicle in 10 Seconds

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Discussion Overview

The discussion revolves around calculating the thrust required to propel a 300lb vehicle a quarter of a mile in 10 seconds. The scope includes theoretical modeling and mathematical reasoning related to thrust, mass, and acceleration in the context of rocket propulsion.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using Newton's second law and proposes a model where the mass of the vehicle changes over time due to propellant consumption.
  • Another participant discusses the integration of acceleration to find velocity and displacement, providing a detailed mathematical approach to derive thrust requirements.
  • There is mention of the importance of specific impulse (Isp) and its effect on thrust, with a reference to typical values for conventional rocket engines.
  • A participant expresses a lack of confidence in their mathematical skills and seeks support for their project, indicating a desire for collaboration.
  • One participant questions the placement of the thread in the Math forum, suggesting it may be more appropriate in a Physics context.

Areas of Agreement / Disagreement

The discussion contains multiple competing views on the approach to calculating thrust, with no consensus reached on the best method or assumptions to use. Participants express differing levels of comfort with the mathematical aspects of the problem.

Contextual Notes

Assumptions made include constant gravitational acceleration and the immediate acquisition of maximum thrust, which may not reflect real-world conditions. The discussion also highlights the dependency on the propellant mass fraction and fuel consumption rate, which are not definitively established.

wolram
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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.
 
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lets begin with Newtons second law
sum F= ma the only problem is m is changing. Let's make one assumption to make this problem a million times simpler. Let's 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 let's integrate acceleration with respect to t. and we obtain v = T/k*(ln(Mo - kt) - ln(Mo)) + Vo, let's 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 let's 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 let's 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
 
more info

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

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