# Calculating Power Requirements for a Ski Tow Rope with Friction | 18.9% Incline

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• rlay023
In summary, this person is trying to figure out how to calculate the power needed to pull a ski tow rope with a weight. If they follow the problem correctly, they need 1.10bhp. They are stumped on how to calculate the friction. There are two possible approaches: 1) adding the normal force (with friction coefficient) and 2) adding the friction force. The second approach is more accurate but requires more knowledge and gear.
rlay023
Hi Everyone -

I found a previous post here that has helped me size an engine for a ski tow rope but wanted to take that a step further an add friction into the equation.

If I followed that problem corrected, the tow rope requires 821W calculated as follows:
power = work/time
work = mhg (272kg)(14m)*(9.8 m/s^2)
time = target speed of 5.5 ft/s, it will take 45.5 seconds to travel 250 feet.

.821kW *. 74569 = .61bHP required

The friction coefficient of snow/ice can range from .03 - .1 and the angle of ascent is 18.9%.

Can anyone help layer in the friction loss into the power calculation?

Welcome to PF.

Can you provide a link to that previous thread? That would help us understand what the context is for your question. Thanks.

Carry your units in all calculations
rlay023 said:
.821kW *. 74569 = .61bHP required
This should be $$\ (0.821 kW)1.34HP/kW=(whatever) HP$$ And choose either meters or feet! Nasa discovered this on mars.

And hurry you only have a couple more good years of snow...

russ_watters
Thanks for your replies. Let's see if I can restate this in all the same units.

What is the required HP to pull 272.2kg up a 14.0M incline with a slope of 18.9% in 45.5 seconds? Using the previous example and the response above, I arrive at 1.10.

Work = mgh = 272.2(9.8)(14.0)
Time = 45.5 second
Power = .82147kW

HP = .82147kW * 1.34 HP/kW = 1.10bHP

The above assumes no friction which we know to be anywhere between .03-.1 depending on the snow/ice conditions. How would we layer the friction loss into the original engine sizing problem?

hutchphd
rlay023 said:
The above assumes no friction which we know to be anywhere between .03-.1 depending on the snow/ice conditions. How would we layer the friction loss into the original engine sizing problem?
Power is force times distance over time. Do you know how to calculate the friction using the skier weights, slope and friction coefficient?

russ_watters said:
Power is force times distance over time. Do you know how to calculate the friction using the skier weights, slope and friction coefficient?
That is where I am stuck. The examples I have found are only on flat surface with friction or slope surface with no friction.

rlay023 said:
That is where I am stuck. The examples I have found are only on flat surface with friction or slope surface with no friction.
The normal force is the cosine of the angle times the weight. That's the part of the weight that acts against the slope.

Something along the line of:

F = uN = .01 * (2619.4258) = 261.94N
where N = Cos(18.9)*272.2*9.8

Power = (F*D)/t = (261.94N)*(75.2m)/45.5sec = 433.10W

The number seems correct (you wrote down ##\mu =.01## but then used ##\mu =0.1## ). You will likely need a gearbox and some rigging which will add some drag and better to have too much power than not enough.. There are many ~6HP engines to be had from China really cheap. Don't know how good they are.

Good catch. I used .1 in the actual equation, as the upper limit of the static friction coeficient - just transcribed it wrong.

Sorry for the basic question here but (assuming the work/power calcs are done correctly on the friction part), how do I factor in the .433k1w from friction. Do I calculate work given the normal force and then add 433.1W to it before converting to HP? Or, do I add it to the initial power calculation given at the beginning of this thread of .82147kw

I would convert the 0.433 kW to HP and add it to the total. There are many ways but since I assume you will be running this on "horsepower" that seems the most useful unit.

Thanks again for all your help.

Depending on the how this is done provides very different answers, which I didn't expect to happen.

Approach 1: If I add .43310kw to the original .82147kw, I get 1.25457kW or 1.68 bHP.

Approach 2: If I take the normal force of 2619.4258N and add the frictional force of 261.942N, I get 2881.36838N times 75.2M (distance) over 45.5 sec (time) = 4.76407 kW or 6.39 bHP.

The sideways distance and the vertical distance are not the same. For gravity you had correctly used the vertical distance 14.4m to get the lifting energy required. For the friction you use the sliding distance 75.2m to get the "sliding work" required. Then you add them. So#2 is wrong .
Also be aware these give average power requirements and will fluctuate if the slope is nonuniform, so leave margin above 1.68 HP.

Awesome. Thanks you so much. I had sized a 5HP to be on the safe side.

hutchphd
In doing further research on the forum, I stumbled upon a torque related post relevant to the same project - https://www.physicsforums.com/threads/help-with-motor-for-pulling-uphill.957569/

Following that as guide, I come up with the following solution:

F = G * sin (theta) + R , where:
G = m(g) = (272.2)(9.8) = 2668.93
sin (theta) = .32
cos (theta) = .95
R = G *cos (theta)*mu = (2668.93)(.95)(.1) = 252.50
F = 2668.93*.32 + 252.50 = 1117.18

P = Fv, where:
v = 1.68 m/sec
P= 1117.18*1.68 = 1872.57

T = P/omega, where:
P = F(v) = 1117.17(1.68) = 1872.57W
rpm = 180.1
omega = v / r = 1.68/.089 = 18.859
T = 1872.57/18.859 =99.30 Nm or 73.242 ft/lb

My first question is whether I selected an applicable approach and secondly did I apply it correctly?

Last edited by a moderator:

## 1. How do you calculate the power requirements for a ski tow rope with friction on an 18.9% incline?

In order to calculate the power requirements for a ski tow rope with friction on an 18.9% incline, you will need to use the formula P = (F x V)/η, where P is power, F is the force required to pull the skier, V is the velocity of the skier, and η is the efficiency of the ski tow system. You will also need to take into account the coefficient of friction between the rope and the pulley system.

## 2. What is the efficiency of a ski tow system?

The efficiency of a ski tow system is the ratio of the work output (or power output) to the work input (or power input). In simpler terms, it is the percentage of the energy put into the system that is actually used to pull the skier up the slope. This can vary depending on factors such as the condition of the rope, the design of the pulley system, and the skill of the operator.

## 3. How does the incline of the slope affect the power requirements for a ski tow rope?

The incline of the slope has a direct impact on the power requirements for a ski tow rope. The steeper the incline, the more power is needed to overcome the force of gravity and pull the skier up the slope. In the case of an 18.9% incline, a significant amount of power will be needed to overcome the steepness of the slope.

## 4. What is the significance of friction in calculating the power requirements for a ski tow rope?

Friction plays a crucial role in determining the power requirements for a ski tow rope. Friction is the force that resists the movement of the rope through the pulley system, and it can significantly decrease the efficiency of the system. Therefore, it is important to take into account the coefficient of friction when calculating the power requirements for a ski tow rope on an 18.9% incline.

## 5. How can the power requirements for a ski tow rope be reduced?

There are a few ways to reduce the power requirements for a ski tow rope on an 18.9% incline. One way is to increase the efficiency of the system by using high-quality ropes and well-designed pulley systems. Another way is to decrease the coefficient of friction by regularly maintaining and lubricating the pulley system. Additionally, having a skilled and experienced operator can also help to reduce the power requirements by using efficient techniques for pulling the skier up the slope.

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