Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General Question on HP required on level ground vs on an incline

  1. Sep 30, 2014 #1

    I been having a "debate" about the load on an engine when going max speed on a lake vs a lesser speed but on an incline..

    Here is the basic jist of the discussion...

    We are talking about a 150HP snowmobile using a CVT at sea level with low snow conditions that is traveling across the lake WOT at max speed (say 110MPH) vs the same snowmobile engine at 9,000ft spinning a much larger track and running in 3ft of snow and on an incline of about 45 degrees WOT traveling at about 45mph up this snow packed hill.

    So, we know that the mountain snowmobile is losing some decent power because of the elevation and we know that it is having to overcome some serious resistance due to the snow depth and the incline..

    How does one determine if these high elevation obstacles out weigh the drag coefficient and higher HP from the snowmobile traveling at 110 MPH on a glass smooth lake?

    Gearing on the lake sled would be in the 2.0 range and gearing on the mountain sled would be in the 2.35 range.

    In other words... which engine is working harder?
    There is probably not enough information to run any true calculations.. if not... would you care to make an educated guess on which engine would be experiencing the most LOAD and STRESS?

    To add another discussion along these same lines.....
    How about this..
    Take these same 2 sleds at place them on a lake at sea level elevation.. BUT keep the longer (163" with 2.5"paddles vs 121" with .75" paddles)BUT gear them the same....

    OK, the sled with the longer and taller track will obviously be slower in terms of top end speed (probably about 15MPH slower) but the HP of the engines and the gearing will be the same.

    Which engine is under a more heavy load when running WOT across the lake? The slower one or the faster one? or would they be under the same load?

    Thanks , in advance, for any consideration anybody gives to this topic.

  2. jcsd
  3. Sep 30, 2014 #2

    Ranger Mike

    User Avatar
    Science Advisor
    Gold Member

    At wide open throttle you have the maximum power you are going to get. This is the max " load , stress" what ever you want to call it. WOT is WOT, period. You will have different results based upon what you do with the power at WOT and this result will vary because of paddle length, gearing and terrain but both engines at WOT will be maxed out.

    The difference between sea level and elevation will impact on the engine if not jetted properly so the advantage would be to the sea level engine if properly jetted for that altitude but not knowing what the jetting is, the advantage may be with the mountain engine...can not tell until one reads the spark plug after WOT test run...
  4. Sep 30, 2014 #3

    jack action

    User Avatar
    Science Advisor
    Gold Member

    I agree with Ranger Mike. That being said, assuming air-fuel mixture is ideal for both cases to maximize power, the snowmobile at sea-level will produce more power at WOT than the one at 9000 ft since the air is denser, hence more oxygen will enter the engine to burn more fuel, giving more power.
  5. Sep 30, 2014 #4
    I agree on power output.. that is not what I am asking.. What I would like to know is which scenario. would the engine be loaded harder? I think that the engine at higher elevation pulling the heavier load would be stress much more than the engine that is at sea level no longer accelerating and at max speed... The main difference is the wind drag of the fast sled vs the lower wind drag of the mountain sled.. But the fact that the mountain sled is trying to do more with less IMO makes it the engine that is under more stress.. what do you guys think?
  6. Sep 30, 2014 #5

    jack action

    User Avatar
    Science Advisor
    Gold Member

    It all depends on your definition of «loaded». A «load» is not a defined physical quantity and is used to define a lot of things depending of context: power, torque, work, weight, ...

    If the power output is the same on both engines (no matter whether the engine is propelling a vehicle at mach 1 or that it «lifts» a vehicle uphill or that it is spinning freely a set of tires, burning rubber on asphalt), it means that the same amount of fuel is burn in the combustion chamber, at the same rate. That produces the same force on the piston and crankshaft, which in turn puts the same stress on the mechanical components.

    The resistance might be different in your cases (wind vs uphill), but the fact that they both require the same power output (WOT @ same rpm), means that they are of the same magnitude.

    I re-read your OP and I noticed that you are not talking about the engines being at the same rpm. If that was the case, the engines would still have the same fuel quantity burn in the combustion chamber (well, it varies slightly due to valve event) for each revolution. This means the same force is applied on the piston, which means the same torque output is expected and the mechanical components are under the same stress (well, actually the faster engine has added stress due to the larger accelerations and decelerations of the reciprocating parts and the increased centripetal acceleration of the rotating parts). But, because there are not the same rpm, less fuel is burn per unit time (less number of revolutions), hence the power is less. So, in this case, one engine is under less stress and less power, but they are still loaded equally (if we assume load is defined as torque).

    As to determine which one is working more, it depends on the time. Work is defined as a force times a displacement. For an engine, it would be the torque applied to the crankshaft multiplied by the number of revolutions. So if the slower engine turns for a longer period of time than the faster one, it is possible that it does more work - even more - if in the end it does more revolutions.

    It is all about definition.
  7. Sep 30, 2014 #6
    Thanks for the reply.

    The engines are BOTH running the same RPM.. they are 2 stroke engines as well.. So the ONLY difference is that one is at a higher elevation making about 25% LESS engine HP than the other.. And the one making the less power is pulling a long hill in deep snow and the one making the most power is on a flat icy lake WOT but no longer accelerating..

    I would think that the lower HP engine tasked with a heavy snow and an incline would be under more "stress" than the one cruising along the lake with the main resistance being the wind drag from the High MPH.. But I could be mistaken and that is why I am asking here to you guys that have more knowledge.

    I think of a Toyota truck vs a Diesel trying to pull the same trailer up hill.. The toyota engine IMO would be under a much heavier stress/load because it is tasked to do more with less.. than the Diesel. Am I off base here?

    Another example would be a boat trying to get on plane vs cruising a WOT Max speed on plane.

    Lastly say a human at sea level dragging a trailer vs that same human at 10,000ft dragging the same trailer.. Ther eis no doubt the lack of air at 10K would make the task more stressful/load than when at sea level

  8. Sep 30, 2014 #7

    jack action

    User Avatar
    Science Advisor
    Gold Member

    As said earlier, same power, then same everything.

    You seem to think that a high speed wind offers less resistance than a snowy inclined trail. But wind can offer great resistance. The reason why you reach a maximum speed, it is because all of your engine power is used to fight the wind resistance and there is none left to accelerate the vehicle anymore, hence no more speed increase. That is how strong the wind can be.

    The Toyota might be «light duty» and might broke down of wear faster than a bigger truck because of its construction, but if both of them pull the same load on the same incline, at the same speed, then they do the same work, with the same power and even the same force.

    If the engine parts of the Toyota are smaller, then of course they will be under a larger stress. Because stress is defined as a force divided by an area. If the area is smaller and the force is kept the same, then the stress increases.

    Again, the high speed wind resistance can be as big as a low speed water resistance.

    Like was said earlier, there is no doubt that the lack of oxygen at higher altitude will result in less maximum power for a given machine, either an engine or a human. Note that I'm talking about maximum power.

    Say an engine can produce 150 hp at sea level and 100 hp at a higher altitude. If you ask the engine to do a task that require 50 hp at sea level or at the higher altitude, the engine is not working harder in either situation. However, assuming they are at the same rpm, the throttle will be more open on the engine at a higher altitude because the air is less dense. Still, the same quantity of oxygen and fuel will be entering the combustion chamber, producing the same power, same everything.

    Now, if the task required to do demands 125 hp, at sea level the engine will do it without even being at WOT. But at a higher altitude, even at WOT, you will simply not be able to do the task. There is no comparison possible since one can't do the job at all.
  9. Sep 30, 2014 #8
    This is great stuff.. I appreciate the information..

    To clarify, the task at hand for the high alt sled is never ending.. meaning, it WILL power out and not finish the climb.. so it will be using ALL the power it has available to it at that time and , yes, it will always be less than the sea level sled..

    I ,now, understand that the wind resistance is a greater force than I had expected.. I just need to be able to know if it is more of a force than the combination of the hill and deep powder.. the other sled is trying to climb.. Again, this hill is not a groomed snow and it can not be conquered with the given sled.. we call this "high marking"

    I think there may be some confusion on what I am calling "load or stress" vs what you are calling "work" So, the scenario is this:

    The high alt engine is giving 100% all it has WOT for as long as it can stay above the snow and not sink.. The sea level sled is done accelerating and is trying to obtain max speed bu can not get any faster (due to wind, drag, gearing, or not enough power)

    I guess what I am struggling with is the "value" or magnitude of the sea level sled's "opposition" vs the same for the high alt sled?

    In my mind.. the hi alt engine is giving it 100% for as long as it can and the sea level sled , while still WOT, is done accelerating and is no longer outputting it max power (100%) and this may be where I am mistaken.. MAYBE, the sea level sled is giving 100% even tho it has finished its acceleration??

    BUT, IMO, the internal engine "load" should be higher on the lower HP sled that is giving absolutely 100% .. I guess this is where the confusion lies.

    Maybe it is just not that simple (cut and dry) as I think...

    So, putting it another way.. Which of these 2 engines (given they ran continuously) in these conditions would give out first???, given they are both mechanically perfect when they began...?? Any thoughts on that?
    Last edited: Sep 30, 2014
  10. Sep 30, 2014 #9
    OH, by the way.. Love the Blackie Lawless Icon...!! One of my favorite bands of all time and highly under-rated!
  11. Sep 30, 2014 #10

    jack action

    User Avatar
    Science Advisor
    Gold Member

    Let's go to basics. To move your vehicle, you need to accelerate it (to change its speed from 0 to whatever). To accelerate it, you need a force. The acceleration will then be equal to that force divided by the mass of the vehicle (Force = mass X acceleration). As the vehicle starts to move, there will be some rolling resistance. As the vehicle speed increases, the aerodynamic drag will increase too. If the vehicle is going uphill, then it also have to fight gravity. These forces all oppose the traction force of the vehicle and, thus, reduce its magnitude. So there is less force left to create acceleration. And since aerodynamic drag increases with speed, at one point the summation of all those resistance forces will be equal to the traction force and there will be none left for acceleration.

    The power needed by the vehicle is the product of the traction force and its speed. So say that at 20 m/s (72 km/h or 45 mph) you know that all resistances (drag, rolling resistance, uphill) require 1000 N (225 lb) of force. That means that 20 X 1000 = 20 000 W (20 kW or 27 hp) of power is required to keep the vehicle at that speed. If you deliver, say, 80 000 W (80 kW or 107 hp) of power instead, then you have 60 000 W of extra power that will go towards acceleration. This will translate into a 60 000 / 20 = 3000 N (674 lb) of additional traction force. If the vehicle has a mass of 500 kg (441 lb), then it will have an acceleration of 3000 / 500 = 6 m/s² (0.6 g's).

    As the vehicle accelerates, if the speed reaches 30 m/s (108 km/h or 67 mph), the resistance might have increased to somewhere around 2000 N (450 lb). Now you need 30 X 2000 = 60 000 W (60 kW or 80 hp) just to keep the vehicle at constant speed. If you are still giving it 80 000 W of power, then you only have 20 000 W of power available for acceleration or 20 000 / 30 = 667 N (150 lb) of additional traction force. So the acceleration is then 667 / 500 = 1.33 m/s² (0.14 g's).

    In the first case the car was going 20 m/s with a 4000 N total traction force and in the second case it was going at 30 m/s with a 2667 N total traction force. In both cases, they were under 80 000 W of power (= 20 X 4000 = 30 * 2667).

    The first graph on this acceleration simulator gives a more complete calculation to evaluate the acceleration at any speed, on a flat track, given a maximum power. When you reach an acceleration of zero, you get the maximum speed. And then it just extrapolates the time and distance from the speed and acceleration to find the 0-100 km/h or the ¼-mile results.

    Again, it doesn't matter what the resistances are, just their magnitudes.

    If they produce the same power at the same rpm and they are physically the same, they will react the same, wear the same and be under the same stresses and thus give out at the same time. Maybe the engine in the slower vehicle will suffer overheating problems due to the lack of wind speed, but that is another story.

    The nickname comes from one of their songs which I always like even though not one of their greatest hits.
  12. Oct 1, 2014 #11
    If they produce the same power at the same rpm and they are physically the same, they will react the same, wear the same and be under the same stresses and thus give out at the same time. Maybe the engine in the slower vehicle will suffer overheating problems due to the lack of wind speed, but that is another story.

    This is the difference.. they make the same power at sea level but the other is at 9K ft so it is making less power.. so you can not use the same HP in the calcs..

    OTHER: All WASP cd's are great IMO.. I prefer "Headless", Crimson, and Unholy over the others but there is not a bad one in the bunch!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook