find value of R

here's a problem i was discussing with my teacher, who told me he'd give me extra credit if i could find an answer. i'm starting to think it's mathematically impossible, though.

1.4R-.4 = R^1.4
what is R?

background:
a regular, four-stroke engine (otto cycle) has this formula for efficiency:
1-(1/compression ratio)^K-1
a four-stroke diesel engine has this formula for efficiency:
1-(1/compression ratio)^K-1*[(R^K-1)/(K(R-1))]
so, given identical compression ratios and K values, is there a value of R that will give these two engines the same efficiency? i'm given compression ratio = 7 and K = 1.4. so, for the same efficiency, [(R^K-1)/(K(R-1))] must equal 1, which i transformed to the original equation. i've been looking at it for a couple hours and kind of gotten stuck, maybe someone can give me some further insight.

www.opera.com

R = 1

But R can't be 1. if R=1, then R^K would be 1, making (R^K-1) = 0. the efficiency of the diesel engine would be 100%.

given compression ratio = 7 and K = 1.4, 1-(1/compression ratio)^K-1 = .54, or 54%. i need to find an R that makes [(R^K-1)/(K(R-1))] = 1 to have the same efficiency from the diesel engine.

>>4
after much pondering, i think i've come a little closer to figuring it out. the R i'm looking for represents the cut-off ratio. a diesel engine works by compressing air as the piston rises, then injecting fuel as the piston goes back down. the fuel is burning as it's injected, so the pressure remains constant as the volume increases. the cut-off ratio is the volume when the fuel injection stops divided by the volume when the fuel injection started. if it is equal to one, that would mean that the volume at the start and end of the combustion is the same, or the combustion is instantaneous, like a regular engine. does this mean that it's impossible to make a diesel engine have the same efficiency as a regular engine, given the same conditions? i'm still struggling to mathematically prove that there's no such cut-off ratio.

Can't you just take the top and set it equal to the bottom (anything divided by itself is one) then solve for R?

Also, at first glance this looks like some differential equations,  but I only vaguely remember this (touched on it in calculus).

>>6
yeah, plugging in K=1.4 and setting (R^K-1) equal to (K(R-1)) is how i got to 1.4R-.4 = R^1.4

the problem is that i keep getting R=1. plug R=1 into the above equations and you get 1^1-1 = 1.4*1-1, which obviously isn't true, since 0 isn't equal to .4

>>7
by the way, if someone can mathematically prove to me that (R^1.4)-1 cannot equal 1.4*(R-1), that would be proof enough that the problem's impossible.

If you want to see all solutions to 1.4R - .4 = R^(1.4), first raise both sides to the power of 5:

(1.4R - .4)^5 = R^7.

Now you have a degree 7 polynomial, so use a polynomial root finder (google) to approximate all solutions.

>>9
what? i've forgotten most of the differential equations i learned. google says the solution is a negative number, a complex number, or the number one. -2.92711 satisfies both versions of the equation, but a negative cut-off ratio is physically impossible, that would mean the volume was greater before combustion. the complex solutions don't really exist, so none of those are possible solutions, right?

>>10
Correct, if none of the solutions to the equation are valid for what you are interpreting them as (e.g. negative cut-off value is invalid), then there are no solutions for your context. Also this is not differential equations, this is algebra and some numerical analysis for approximating roots.

Setting the efficiency equations equal to each other gives K(R-1) = R^(K-1). Using Mathematica to solve the equation with K=1.4, it's R=1.92897. There probably is an analytical solution but who cares, maybe try using log on both sides.

>>12
It's R^K - 1, not R^(K - 1).

>>13
o ok. r=1 is the only solution then.

1-(1/compression ratio)^K-1 = E

1-(1/compression ratio)^K-1*[(R^K-1)/(K(R-1))] = E

1-(1/compression ratio)^K-1 = 1-(1/compression ratio)^K-1*[(R^K-1)/(K(R-1))]

(1/compression ratio)^K-1 = (1/compression ratio)^K-1*[(R^K-1)/(K(R-1))]

1 = [(R^K-1)/(K(R-1))]

R^K - 1 = K(R-1)

R = 1.

>>1
MAPLE AND MATHEMATICA MOTHERFUCKER DO YOU USE THEM

they cost like $1000

>>16
>>17
i solved it on my quaint old ti-83plus immediately, i just wanted some support for my proof. i think i'm pretty solid in my answer now. thanks for the help, guys.

>>17
Solution #1: Warez
Solution #2: Go to a decent university that gives out student software license.

just use maxima

I feel the need, the need for weed!

Marijuana MUST be legalized.