# Power needed to climb constant slope at constant speed

by daveyjones97
Tags: climb, constant, power, slope, speed
P: 777
 Quote by simurq I should have mentioned that by "vehicle" I meant an aircraft. Unlike a car, an aircraft must cope with effects of drag and gravity more than a car mostly subject to ground friction. Therefore, energy saving and particularly the notion of "excess power" is of crucial importance for any aircraft regardless its mode of flight - climb or level. Hence my question - is there a formula or approach to determine the relationship between (i) the rate of change of power with altitude; (ii) decrease in speed and climb rate; and (iii) effects of other factors such as air temp, pressure and density. PS: Btw, did you take a look at attachment from my previous post? It will definitely help you to visualize my point here.
I understand what you're saying and the concept of "excess power" (power beyond what is necessary for level flight). I believe perhaps you used the terms "power" and "excess power" interchangeably. This being a physics forum, the term "power" has a specific meaning. Basic physics for a car, plane, rocket are the same: a steady climb should require a steady power (unless some variable changes).

I saw the attachment, and indeed it appears that the plane loses power as it climbs, and I still suspect it is direclty related to decreasing air pressure. For an air breathing engine, less air means less power. Nevertheless, the power required to maintain the climb remains the same (it actually probably decreases). But since the "excess power" of the plane decreases, so does it's rate of climb. It's engine is getting weaker, as it's simply running out of breath.

I don't know the formula for this but I'm guessing the air pressure is the driving factor.
P: 777
 Quote by simurq 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?!?
I went back to this post and I would like to point out the inconsistancy, as I believe my above post might be confusing.

 Logically, any vehicle (and aircraft, in particular) climbing up the hill is doomed to lose power directly proportional to the height of climb
As sophiecentaur said this might be true due to practical reasons (less air), but it's not nesessarily so as it's not the result of some fundamental physical law.

 In other words, the more a vehicle climbs, the more excess power it needs at his disposal.
This is NOT true. To maintain the climb, it needs the same power, and the same excess power. What is preventing the plane from maintaining the climb, is that it simply does NOT have the same excess power. Its excess power is decreasing with altitute: the engine is getting weaker. The necessary power to maintain the climb, however, remains the same.
Thanks
PF Gold
P: 12,256
 Quote by Lsos I went back to this post and I would like to point out the inconsistancy, as I believe my above post might be confusing. As sophiecentaur said this might be true due to practical reasons (less air), but it's not nesessarily so as it's not the result of some fundamental physical law. This is NOT true. To maintain the climb, it needs the same power, and the same excess power. What is preventing the plane from maintaining the climb, is that it simply does NOT have the same excess power. Its excess power is decreasing with altitute: the engine is getting weaker. The necessary power to maintain the climb, however, remains the same.
There are too many variables involved here to make a definite statement about the situation. If a vehicle (car) is driving up an incline and is totally self powered, the energy needed per metre of height is constant until g reduces significantly (and that is only at immense heights - do the 1/rsquared sums). At a given speed, the drag would be less as height increases.
For an aircraft which is using the air for lift then its design will be optimal for a certain air density. This could favour a very high altitude (as with the U2 spy planes) so the power needed to climb may be less at high altitude. Then the characteristics of a 'real' engine are almost bound to involve loss of power beyond a certain height. This may not actually be relevant to the original question - if the question is to be answered literally.

I think the question needs to be broken down into parts, each of which stands a chance of being answered.
 Sci Advisor P: 2,470 Lsos is absolutely correct. If you define excess power as power over or under power required to maintain speed at fixed altitude or on level ground, then constant rate of climb will require constant excess power. While the power requirement will change with altitude, the excess power requirement will remain the same.
P: 777
 Quote by sophiecentaur There are too many variables involved here to make a definite statement about the situation.
I agreee to an extent (I even mentioned that power requirement may fall due to diminishing g and diminishing air resistance). However, I brushed these variables aside and simplified the situation based on the following statement by OP:

 Logically, any vehicle (and aircraft, in particular) climbing up the hill is doomed to lose power directly proportional to the height of climb.
which led me to believe that OP thinks there is some intrinsic property of gravity which tends to pull down an object harder the higher we go, when in fact the invere is true. I primarily wanted to get that out of the way and not cloud the issue unnecessarily.

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