What is the Maximum Gradient a Vehicle Can Climb Based on Known Load Conditions?

In summary: Weight of vehicle : 1800kgWeight of payload : 500kgMax gross weight of vehicle : 2300kgSpeed of vehicle at start of gradient : 50kphRoad condition : Dry, paved road in good conditionCalculation 1 : Max available power : 120kw / 160hp Maximum Gradient : 12%
  • #1
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I need to calculate the maximum gradient that a specific vehicle will be able to climb when subjected to known load conditions. There are 6 calculations required for 4 sets of conditions.

Values common to all calculations: -

Weight of vehicle : 1800kg
Weight of payload : 500kg
Max gross weight of vehicle : 2300kg
Speed of vehicle at start of gradient : 50kph
Road condition : Dry, paved road in good condition

Calculation 1 : Max available power : 87kw / 116hp Maximum Gradient : ?

Calculation 2 : Max available power : 120kw / 160hp Maximum Gradient : ?For calculations 3 and 4 the common values remain the same but the vehicle is now towing a trailer of 750kg

Calculation 3 : Max available power : 87kw / 116hp Maximum Gradient : ?

Calculation 4 : Max available power : 120kw / 160hp Maximum Gradient : ?For calculations 5 and 6 the common values remain the same but the vehicle is now towing a trailer of 1500kg

Calculation 5 : Max available power : 87kw / 116hp Maximum Gradient : ?

Calculation 6 : Max available power : 120kw / 160hp Maximum Gradient : ?

I would be grateful if someone could instruct me as to how to calculate the maximum gradient which would then enable me to calculate the value under a wide variety of conditions.

Thanks in advance for any help provided.

T.

Mod note: Thread is not homework. Moved to engineering.
 
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  • #2
You also need the stall speed for the engine.
Given the power and the speed you can get the rate of climb... which tells you the gradient.
 
  • #3
Hi Simon, thanks for your reply. Apologies, I forgot to mention the stall speed.

Theoretically the stall speed of these vehicles is Zero as they are 4x4 Petrol/Electric Hybrids. When the 4x4 Lock is activated each wheel is driven independently by the electric motors only. The petrol engine engages directly through a transaxle which means that it can only engage above a certain minimum speed as there is no gearbox - and that minimum speed is 70kph. That is why I set a condition of "50kph at the start of the gradient" so as to eliminate the petrol engine. If I set a condition of "80kph at the start of the gradient with full throttle" then the power available from the petrol engine could be added to the equation with a "decoupling speed" of 70kph replacing the "stall speed".

My problem is that I simply don't have the maths knowledge to know how to even start to solve such a problem. If you could give me the method to the solution then I can slot in the various numbers to arrive at a solution for each given set of conditions.

Once I know how to solve the problem then I am hoping to set up an Excel SS which will enable me to quickly calculate the results for a set or parameters.

Thanks for your time.

T.
 
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  • #4
OK - so the vehicles can, in principle, go arbitrarily slowly?
So the limiting factor is the grip of the wheels on the surface.

Consider - the work done to go up the ramp is the gravitational potential energy gained.
This energy is suppled by the motor.

If the vehicle travels along the ramp, angle \theta ot the horizontal, at constant speed v, then the engine is supplying ##P=mgv\sin\theta## power just to go up the slope.

Presumably there is some power consumption just in normal operation, traveling in a straight line, at constant speed, on level terrain.

You can kinda see that if you have a fixed power, then a steeper slope is ascended more slowly ... but there is no maximum gradient that can, in principle, be ascended. But if too steep - the wheels will slide.

In practice there is a maximum load a motor can take without something going wrong.
An ICE will stall, an electric motor can burn out or just jam.
 
  • #5
I think that I am starting to get a grasp of this... maybe :)

It is safe to assume that the manufacturer of the EV will have incorporated some kind of cutout override if the very expensive electric motor was on the point of self-destruct. In a practical sense one could therefore think of that EV cutout-point to be equivalent to the stall-point of an ICE.

About 5 minutes ago I uncovered an interesting piece of info published by the manufecturer of these vehicles. Having digested reams of technical data, deep within this considerable pile I found a 1-line reference to my question :-

"... towing capacity of 1,500kg at a maximum 12% gradient... "

Hmmm... and of course in my mind that raises another question - How did they calculate the 12% :confused:

T.
 
  • #6
That's what you are looking for.
They'd have done that by a mixture of measuring and conservation of energy calculation.
Note: 12.5% is a 1/8 gradient, so they probably have a ramp that they test new EVs on.

That figure is a "safe figure" rather than a maximum possible. Best not to exceed that.

i.e. do not exceed 1500kg, and do not exceed 12% gradient.

Have you tried the EVs on different terrain?

For angles less than 20deg, if you double the angle you roughly double the rate that work is being done at the same speed - so you want to halve the total mass or halve the speed to compensate.

Off the figures, I imagine the EVs are pretty poor hill climbers. i.e. there are some slopes so steep that they just sit there and strain.

For a random vehicle it may be that a steep slope just leads to skidding - the wheels turn but there is not enough traction. Some vehicles will be powerful enough to climb any slope they can get purchase on.The basic calculation is that the energy the EV has at the bottom of the hill, plus whatever it can generate in it's power plant, turns into gravitational PE and kinetic energy (with heat, sound, work against friction) as it climbs the hill.

So if it starts at the bottom of the slope with speed u, and the slope makes angle A to the horizontal, then:

##\frac{1}{2}Mu^2 + Pt = \frac{1}{2}Mv^2 + Mgv\sin A + W\!\!_ft##

M is the combined mass, Wf is the power loss due to friction, and v is a function of time.
Wf may depend on the load... you'll see this if the max speed changes with load.

We add in the proviso that the vehicle does not skid on the slope so:

##\mu\cos A > \sin A## ... where mu is the coefficient of static friction.

[edit: something like - ##P-W_\!\!f < \mu Mgv\cos A##]

If the grade is tackled at a constant speed, then v=u, simplifying the equation.
If the rolling resistance loss is very small compared with the other terms, then you may be able to put Wf=0 too.
But I suspect not.

BTW: if the manufacturer has not said (i.e. in the manual) there there is a safety cutout to avoid burning out the motors under high loads, do not assume there is one. There probably is, but if you exceed the manufacturers specifications and something breaks, they are unlikely to accept liability - and it could impact on your insurance cover.
 
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  • #7
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Hi Simon

Firstly, my apologies for a delayed response - I was away from my desk all day yesterday.

Secondly, a really BIG THANK YOU for such an excellent reply. And from my perspective, the best part about your reply is that you have actually taught me something. Learning somethng new every day is one of my "things-in-life" ... and what I have learned from your reply certainly fulfills that goal for today :)

Away from the theory for a moment and looking at this particular vehicle in the real-world, a very interesting result, not from a professional tester but just from a group of "adventurers", is that last winter in particularly heavy snow and icy conditions this vehicle outperformed every other vehicle in the group which included all the "Big-Bully-Boys" of the offroad world. Where the BBB's were all getting stuck in snowdrifts and slip-sliding-away on the icy surfaces, this vehicle performed as if it was on a paved road. This is largely due to the smooth torque-curve that is an inherent property of an electric motor. And when you have 4 wheels each being independently driven by electric motors then it really is no-contest with the more conventional drivetrain.

In addition to outperforming the BBB's, when the BBB's were all belching out clouds of smelly fumes, frightening the wildlife with the noise and generally being about as environmentally unfriendly as it is possible to be; this vehicle, because it is a Hybrid and is in full EV mode when in 4x4 lock produces zero pollution and is virtually silent - even the artic foxes sat back and applauded us when we went by :)

But now to polish up my Excel skills and see if I can take what you have taught me and apply it to an SS in a practical sense so that I know the limits to which we can push these vehicles.

Thanks again, you have been a great help.

T.
 

What is the maximum gradient question?

The maximum gradient question is a scientific concept that refers to the steepest possible change in a physical quantity. This can be applied to various fields such as mathematics, physics, and engineering.

Why is the maximum gradient question important?

The maximum gradient question is important because it allows scientists to understand the limits and boundaries of physical phenomena. It also helps in designing experiments and predicting outcomes in various fields of study.

How is the maximum gradient question calculated?

The maximum gradient question is calculated by finding the maximum value of the derivative of a function. This involves taking the derivative of the function and setting it equal to zero to find the critical points, and then evaluating the function at those points to determine the maximum value.

What are some real-life examples of the maximum gradient question?

Some real-life examples of the maximum gradient question include determining the steepest slope of a mountain, finding the maximum rate of change of temperature in a chemical reaction, and calculating the maximum acceleration of a moving object.

How does the maximum gradient question relate to optimization?

The maximum gradient question is closely related to optimization because it involves finding the maximum value of a function, which is often the goal of optimization problems. By finding the maximum gradient, scientists can determine the optimal conditions for a particular system or process.

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