
#1
Dec911, 11:36 AM

P: 14

i failed science at school and often want answers to q's thank God for pf!
i want a nice easy way to work out how much power you would need to climb hills: power versus weight and gradient is what im after. assume constant speed constant gradient ignore wind resistance. im after simple ball park calculations. nothing i've seen so far is quite what im after and im not too hot on algebra. any answers in terms a 5 year old would understand please. 



#2
Dec911, 12:00 PM

P: 617

It depends on how efficient your mode of transportation is (your legs, bicycle, truck) on a certain surface (wet, sandy, dry). Assuming perfect efficiency, all you have to do is overcome gravity. If your weight is w, and the hill has a height of h, then it will take a total amount of energy wh to reach the top of the hill. The rate at which you expend energy to reach the top, or the power, depends on the time you take. If you go on a windy trail, it takes less power but a longer time so you expend the same energy in the end. The power would be
P = wh/t where t is the total time to reach the top. Let's say you don't know the time it takes you, but you know your average speed v and the length of the trail to the top x, then from v = x/t, the power needed becomes: P = whv/x Let's say you don't know the height of the hill, but do know the average steepness of the trail, measured as an angle θ in degrees from a flat trail. From sin(θ) = h/x we have: P = wv sin(θ) where P is the power required (or energy expended per second), w is your weight (not to be confused with mass), v is your total average speed, and θ is the angle of your path with respect to the flat ground. Again, this equation assumes that your system is perfectly efficient (no internal losses, or loss to friction), and you are able to maintain a constant speed v. 



#3
Dec911, 12:16 PM

P: 14

im fine with assuming 100% efficiency, so 130kg x 16kph x 20 degrees = power? 41600, joules? i swear im not deliberately being dumb! im the poster boy for those who spent school looking out the window only to need to relearn it all later in life. if im right the end result is in joules?




#4
Dec911, 12:18 PM

P: 617

power needed to climb constant slope at constant speed
For instance, if you weigh 200 pounds (in SI units that is w = 890 kg m/s^{2}) and the road inclination angle is θ = 10°, then:
power in Joules/second = 154 times the speed in meters/second 



#5
Dec911, 12:19 PM

P: 617

You have to make sure you have everything in the same units system.




#6
Dec911, 12:33 PM

P: 14

i apologise for having such a base level of knowledge, luckily i have the sense to use the same measurement system, and to check that if i get a silly result. 200 pounds is 90.7kg and m/s squared i thought was acceleration. in an effort to help myself im looking for just a constant speed so dont understand 890kg m/s unless thats a watt hours alternative/equivalent measure?




#7
Dec911, 12:41 PM

P: 617

Weight and mass are not the same thing. Weight is mass times the acceleration due to gravity: w = mg. Pounds is a measurement of weight, whereas kilograms is a measure of mass. Let me rewrite the equation:
P = mgv sin(θ) where P is measured in Joules per second, m is measured in kilograms, g is the acceleration due to gravity, g = 9.8 m/s^{2}, and v is measured in meters per second. 



#8
Dec911, 12:55 PM

P: 14

thanks, think i must have been dropped on my head as a baby... i was aware of the weight/mass difference but not pounds / kg. think im having a bad day, any chance you could do an example equation with numbers that i can follow from start to finish? i dont want to waste your time for too long. is there an ebook link i could bookmark for future?




#9
Dec911, 02:49 PM

P: 14

got it i think, its the units that im unsure of but:
90kg x 9.8ms gravity x 16 kph x 10 degrees = power needed of 141,120 joule/min or 2352 watts. my issue was 141120 looked ridiculous to me and i had to work out what i was missing. hopefully im right thanks to chrisbaird for not flaming me! 



#10
Dec911, 06:16 PM

P: 1,351

You need to use the sine of 10 degrees, not 10. sin(10) = 0.1736... this means that going up a 10 degree slope will take about 17% of the power of going straight up with the same speed. You need the speed expressed in meters per second, since the acceleration of gravity is also expressed in meters/second. 16 km/h is (16/3.6) m/s you had no reason to use a unit of joule/min for your answer, since minutes were not used in your units for mass, gravity or speed. Actually this is an achievement not much faster then what a professional cyclist can achieve over an hour long climb. (something like 450w for 80 kg including bicycle), so your answer of 2352 W is still much too high. 



#11
Dec1011, 04:56 AM

P: 14

so 90kg x 9.8ms x 4.4ms x 0.17 (sin10deg) = 659.7 joule sec which = 659.7 watts.
its the combination of unfamiliar units and algebra not to mention the silly mistake not converting the kph that was the problem. i was pretty sure 2000+ watts was too much but couldnt see wood for the trees. its only my hobbies that ever bring me near a calculator. i havent studied for 20 years, as you can tell! Thanks and thanks for your patience. 



#12
Dec912, 12:32 PM

P: 5

I've come across this discussion while asking similar questions myself re: aircraft climb performance. Pls see attachment at my thread above to visualize the problem.
So, judging from your responses and assuming nil winds @ standard ISA conditions (t = 15°C; ρ = 1.225 kg/m^{3}), the power required to climb an aircraft from sealevel to 2,000' will be: Engine's rated power: 230 BHP (~172 kW) Aircraft's weight: 2,950 lbs (1,338.1 kg) V_{y} (bestrateofclimb speed): 78 knots (40.13 m/s) Climb rate @ V_{y}: 1010 fpm Climb gradient: 1010 fpm / 78 knots = 12.95% sin(θ): sin(12.95) = 0.224079
Thank you! Rustam 



#13
Dec912, 12:39 PM

Sci Advisor
PF Gold
P: 11,352

This looks a bit odd, though 



#14
Dec912, 01:02 PM

P: 5

Correct, thanks for the hint, Sir! But the result is not that different... However:
1010 fpm = 5.13 m/s 78 knots = 40.13 m/s Climb gradient = (5.13 / 40.13) x 100 = 12.79% sin(θ) = sin(12.79) = 0.221301 Therefore,




#15
Dec1012, 03:24 AM

P: 5

Although I'm satisfied with above results, the fundamental question I'd like to ask is:
What is the rate of change of POWER with climb? Logically, any vehicle (and aircraft, in particular) climbing up the hill is doomed to lose power directly proportional to the height of climb. And at some point in time it will just stop climbing! In other words, the more a vehicle climbs, the more excess power it needs at his disposal. This can be clearly seen on the attachment as well  the higher the aircraft climbs, the less is the speed, climb rate (nose pitch and angle of attack) and engine performance (although the latter is not visible on the chart, it can be deduced applying the same formula in above posts to different altitudes). So, is there any formula to calculate the rate of change of power keeping the relationships with speed, climb rate and other factors (like temp, density, etc), if any?!? Thank you! Rustam 



#16
Dec1012, 04:27 AM

P: 768

I don't see why you would necessarily lose power when climbing. For airbreathing engines, indeed the higher you climb the less air (and hence power) you have, but this will also vary depending on what type of engine you have and isn't a necessary constraint. So the power requirement for climbing should remain the same, but the means of producing this power CAN diminish.
Actually, power requirements for climbing can even decrease if you take into account diminishing gravity... 



#17
Dec1012, 10:48 AM

Sci Advisor
PF Gold
P: 11,352

Are we confusing the terms Power and Energy here? Some of the statements could use the words interchangeably and mean different things.
There are practical reasons why a normal IC engine might lose power at height but that wouldn't apply to electrical propulsion or an IC engine with its own oxygen supply. 



#18
Dec1012, 12:35 PM

P: 5

PS: Btw, did you take a look at attachment from my previous post? It will definitely help you to visualize my point here. These pages are very good but since they're aimed at pilots (who usually use flight computers to concentrate on a flight rather than wasting time with complex calculations), a lack of thorough technical discussion is obvious and pretty welcome. Rustam 


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