Sanity check: calculating cost of transport in KSP

[seanth]

I've been thinking about cost of transport a lot recently(see http://en.wikipedia.org/wiki/Cost_of_transport and http://www.neodymics.com/Images/EPPaper080229E.pdf). I was hoping someone could sanity check this for me:

If COT=P/(m*g*v), where P=power (joules/s, where a joule(J) is N*meter), m=mass(kg), g=acceleration of gravity (9.81m/s^2), and v=velocity(m/s)

And P=0.5*F*Isp*g, where F= force (in kN), Isp is the specific impulse of an engine, and g is the acceleration of gravity (9.81m/s^2)

Then if you used two basic jet engines to power a kerbal car that weighed 1 metric ton(=1000kg) and reached a maximum sustained velocity of 50m/s, and using thrust (100kN)and Isp(7282s. I think Isp in KSP is by weight?) values for those jet engines, the COT calculation would be (sorry for showing all the steps):

COT=P/(m*g*v)

COT=2*(0.5*F*Isp*g)/(m*g*v)

COT=2*(0.5*100kN*7282*9.81m/s^2)/(1000kg*9.81m/s^2*50m/s)

COT=(100000N*7282s*9.81m/s^2)/(9810N*50m/s)

COT=(981000J/s^2*7282s)/(9810N*50m/s)

COT=(7143642000J/s)/(490500J/s)

COT=14564(unitless)

Am I doing anything supremely dumb here?

Edited October 3, 2013 by seanth

[samstarman5]

As someone who only got as far as thinking of taking on Trigonometry and then Calculus, your work sure as heck doesn't look dumb to me. Bear in mind that there are many factors that real life has to deal with that KSP does not. Things such as lack of variances in gravity by mass, secondary and tertiary gravity influences(secondary being the Moon on an Earth satellite, tertiary being Jupiter on that same satellite as examples which even affect us on the ground). That just serves as an example of the many differences that exist.

The fact you are willing to make such effort shows you are nowhere being dumb, and even if your math and figuring is off, I bet it ain't by much. Hell, back in the days of the Phoenicians and Egyptians, one man figured the circumference of the Earth with an error range of mere meters just by looking at the shadows of two obelisks on two separate ends of a kingdom the size of Utah.

If there is one thing I conclude from all this, is that we need MOAR BOOSTERS! Cheers!

[adinfinitum]

Just looking through what you've got there, I was wondering where you got P=0.5*F*Isp*g? Working through the units gives me something weird for units of power, I think what I got is kg*m^2/s^3.

[Captain Sierra]

Just looking through what you've got there, I was wondering where you got P=0.5*F*Isp*g? Working through the units gives me something weird for units of power, I think what I got is kg*m^2/s^3.

Could that not be expressed as: kg(m/s²)²? Wait,I don't think it can. Disregard for now, let me consult some physics and calculus textbooks and get back to you on that.

EDIT: no it definitely cannot, that would come out to be m/s^4

EDIT II:

COT=(981000J/s^2*7282s)/(9810N*50m/s)

COT=(7143642000J/s)/(490500J/s)

You are dropping another second here. The first one should be J/s². Which gives you watts squared times watts.

That comes out as watts to the third, or J/s/s/s, which seems off. there is one too many s's in there. Watts per second would be exponentially increasing power. Not sure what watts per second per second is as watts is already a time-dependent unit (J/s)

Edited October 3, 2013 by Captain Sierra

[fatfluffycat]

Too....much....math.....*dies*

[adinfinitum]

With wikipedia as my source, it looks like power is the integral of force times velocity, with respect to time, if you are traveling in a straight line. If force and velocity are constant then it just comes out to power being force times velocity.

With that in mind, lets try it this way.

COT = p/(m*g*v); p = F*v, so...

COT = (F*v)/(m*g*v); The velocities then cancel out and...

COT = F/(m*g);

Looking at this, it seems that the equation has come out to thrust divided by weight, which is interesting, and I hope I didn't do anything wrong. Let's continue on...

The thrust is in kilonewtons, so we need to convert to newtons, just multiply by a thousand...

Now

COT = 100,000N/(1000Kg*9.81m/s^2)

Which gives us a unitless Cost of Transport of 10.2.

Let me know if I did anything wrong in that, having the velocities cancel out that easily makes me a little wary of my reasoning.

[Odo]

Yes, erm, your math looks perfect.

[adinfinitum]

I was just doing some thinking as I was cleaning a few dishes, and I practically typed it out but didn't even realize what it meant as I typed it up. In this simplified case of constant velocity and thrust in a straight line, the equation reduced to the thrust-to-weight ratio of your vehicle.

[samstarman5]

I was just doing some thinking as I was cleaning a few dishes, and I practically typed it out but didn't even realize what it meant as I typed it up. In this simplified case of constant velocity and thrust in a straight line, the equation reduced to the thrust-to-weight ratio of your vehicle.

Nice thing about washing dishes, leaves a lot of brain time free. I wonder if Von Braun's designs came up over some squeaky clean china?