In the gym where I do my in person client training, we have six treadmills, but only one goes above 10 miles per hour.

After regularly training people for six years, I’ve had very few people where this would be a concern. But one of my clients (who is 59 years old!) has been exceeding 10 mph for a few months now. (For quarter mile runs. We’re starting to seriously consider some masters events. Maybe the 800 meters to start.)

I’ve seen some runners complain about this with their treadmills as well. I’ve even seen some joke how their treadmill doesn’t go fast enough to be the average speed for top marathon runners, where your treadmill needs to go above 12.5mph. (Marathon runners are doing what is a sprint for many!)

Or if you’re looking to get some sprint work on a treadmill, and going outside isn’t feasible for some reason.

The one treadmill we have access to that does go above 10.0 mph is ironically the oldest treadmill in the gym. I’ve been fearful the thing would stop working one day. Or, what if someone else is on it at the exact moment we need to be? How could we replicate that amount of output?

A common way to make the treadmill harder is to bump the incline. So I started thinking, how much of an incline would you need to equal the output? If we want the output to be 10.5 mph, but the treadmill doesn’t go above 10mph, how much of an incline could we use to make things similar?

Teacher “Come on everyone! Look alive!”

12 year old “Ugh, what am I learning this math for? Am I ever going to use this? I don’t get why this is important?”

[waiting for the response of “Because I’m an adult and I say so…]

Teacher “When you’re older and over the hill, you are going to be using a treadmill, to be in shape and not be still. One day, along the way, you’ll come across one which moves like poo, what will you do? You’ll be thinking, ‘is my workout feasible??? Why must all hotel gyms be so terrible?’ You’ll be in a rough neighborhood because you never joined the good student crew, and, not wanting to hear the rapist go “boo!” run outside you won’t want to do. You’ll remember this class, get a tickle in your ass, “oh! I can calculate this…but I didn’t pay attention that day, piss.’ You’ll then be more susceptible to the sleaze of heart disease, from not being able to burn off that cheese.

This will help you feel like a loser, all in all causing you to die sooner #TooManySchooners. The opposite sex will be less likely to want to play, because you appear to move like jelly all day. The jelly exercise could have hid, now it’s less likely you’ll have a kid.

To avoid this wrath, you could just learn this math.”

–

### Preface to the model

From a theoretical perspective, I’m not sure how accurate this post is. My brother Chris, who helped a ton with this post, and is about to finish up his physics degree in his junior year (he’s pretty decent at physics), isn’t sure either. Nor is the physics grad student he talked to, or his multiple STEM roommates. I also had a brief discussion with a high school physics teacher, and a rocket scientist about this. The rocket scientist’s response, “I’d probably just put them on the treadmill and find out.”

1) If you think you have a better / more accurate way of deducing all this, please let us know. My brother and I would be very interested!

2) The numbers that end up coming out of this model strike my practical side as “that’s about right,” along with agreeing with actual client numbers. I think it can provide a pretty solid starting point should you find yourself in this situation.

I’m also going to give the equations before giving the reasoning behind them. This way for those where physics is foreign to them, you hopefully won’t feel overwhelmed. I made it so you can literally click a link to the equation, type in a few small numbers, and you’re done. I also give a graph you can read off of too, so you might not have to type anything in.

#### –

### Why not just put them on the treadmill and find out?

This suggestion was tossed around a couple times, “you should probably just do an experiment,” and it’s where the physics / math people don’t have the exercise / training experience. The reason to go through this trouble is, particularly with a higher level athlete (but really anyone!), you don’t want to give them a workload demonstrably above what they’re used to.

The client I have who was the impetus for all this can currently run 10.6 mph on flat ground. I wouldn’t want to put them at a speed and incline that gives a demonstrably higher workload than 10.6 at no incline. (For a higher level athlete, a percent or two can mean “demonstrably higher.”) Sure, I could be cautious and start them low, but then we potentially waste time by ramping up and trying to see where that threshold is. It’s not terrible, but we could be more efficient.

It’s like knowing someone’s bench press max. You don’t go, “alright, we know you can do 315lbs for 10 reps. Let’s see what you can do for 1 rep. Ah, let’s just put 500 lbs on the bar and see what happens.” Instead we have various charts that have been derived to give a **starting point** for what a person’s one rep max is, based on their 10 rep max.

And it’s better than saying, “Alright, let’s put 320lbs on the bar and see how one rep goes. Then 325.” etc. You want to have a good feel for that starting point, if you can.

Start too high and you increase the odds of hurting someone. Sometimes, this doesn’t have to be that high either. If someone’s bench press max is 350lbs, but you start at 375lbs, you’re pushing someone into a realm they’ve never been. Even if you only give them 355lbs, the person will still be pushed into an exercise where they will fail. You’re playing with fire in that case, and you *really* don’t want to fail on a treadmill. As much as possible, you don’t want to overshoot someone.

So, the idea with all this, is if I did put that client on an incline, 5 degrees is about the max I’d start with if they weren’t used to an incline (discussed later), I’d want to be sure I put them at a speed that gave a workload right in the neighborhood of what they’re used to, if not less. And then hopefully only have to marginally work our way *up *from there if needed. You don’t want to have to work your way down, if you can avoid it. And if you are working your way down, hopefully it’s not by much.

This is a longer way of saying whatever value(s) you come up with from below, I’d still start out perhaps 1-3% lower than that number. I give further rationale for this in the caveats section.

–

### The equations

I’m going to start with the equations for those who aren’t interested where things come from. The first equation is what most think is theoretically best, but practically, all the equations come out with similar numbers.

You can simply copy and paste the equation into something like google or WolframAlpha (I’ve already done this for you), substitute the pertinent variables, and get your values.

*Note the units my American friends, and I’d still scroll down and read the caveats section, so you don’t get too lost using this model with someone that perhaps doesn’t fit it.*

Each equation is also a link. Click the link and the equation will already be filled out for you in WolframAlpha. Substitute actual numbers in the for the variables, and you’re all set.

Variables:

V & V1 = speed at no incline (meters / second)

V2 = speed at incline (meters / second)

Sometimes you are calculating this. If you’re trying to calculate the incline angle though, you’ll already know this. This will be made more clear in the examples.

m = mass (kilograms)

g = 9.8 m/s^2 (acceleration due to gravity)

angle = treadmill angle (degrees)

0 to 15 degrees, which is the max treadmill height (at least I’ve never seen one go higher)

d = stride length (meters)

1.5 meters should work for most.

–

Equations:

**V2 = 2.2369 * (( 1/2 * m * v^2 * 1 / d ) * ( v )) / ( 1/2 * m * v^2 * 1 / d + m * 9.8 sin(angle))**

Example: A 54kg person, running at 4.47 m/s (10mph) with no incline, with a stride length of 1.5 meters, wants to know what speed they should be at when running at an incline of 5 degrees?

-> 119 pounds / 2.2 = 54 kilograms.

10 miles per hour / 2.24 = 4.47 m/s.

2.2 is the conversion factor for lbs to kg; 2.24 for mph to m/s.

Click the equation, enter the numbers, and,

–

-> This is if you want a graph:

–

**angle = sin^-1(((( F * V1) / V2) – F ) / (m * 9.8))**

Where **F = (1/2 * m * V1 ^2) / d**

V1 = speed at no incline

V2 = speed at incline

Example: A 54kg person, with a stride length of 1.5 meters, who can run at 4.47 m/s (10mph) on flat ground wants to run at 4.9 m/s (11.0mph) on flat ground, but their treadmill only goes to 10mph. If they bump the incline and keep the speed at 10 mph, what incline should they be at for things to be comparable?

F = (1/2 * 54 * 4.47^2) / 1.5

= 398

angle = sin^-1(((( 398 * 4.7) / 4.47) – 398 ) / (54 * 9.8))

= 2.12 degrees

–

**V2 = [(V1^2) / (V1 + g*Sin(angle))] * 2.24**

Example: A person who is running at 4.47 m/s bumps the incline to 5 degrees. What speed should they be at for things to be comparable?

V2 = [(V1^2) / (V1 + gSin(angle))] * 2.24

V2 = [(4.47^2 / (4.47 + 9.8*sin(5))] * 2.24

= 8.4 mph

(make sure you use degrees here, not radians!)

–

### The first way

When we run on a treadmill we exert a force on that treadmill, which we do over a certain amount of time. In other words, we are generating power.

Power = Force * Velocity

When on an incline, we are doing the same thing, but we are having to overcome another force, which is the incline.

*If you want to know where the above comes from, try here. *

What we’re primarily concerned with is the extra force from the incline. The vertical forces are going to cancel one another out. (Further addressed in the “Caveats” section.)

When we add that extra parallel force to our original no incline force, we have our force on an incline at,

Fincline = F + mgSin(angle)

(where F is our original F)

Meaning our power at an incline is,

Power = (F + mgSin(angle)) * V2

Our two equations right now,

P = F * V1

P = (F + mgSin(angle)) * V2

V1 is the velocity we ran at with no incline.

V2 is the velocity we run at with an incline.

What we want to know here is what is V2? What velocity do we need to run at with an incline, in order to equal the same power we run at with no incline?

For instance, say we want to know how much power does running at 11.0 mph generate? But our treadmill doesn’t go up to 11.0mph. It only goes up to 10.0mph. What incline do we need to run at, at 10.0mph, in order to equal the power of no incline at 11.0mph?

We set our equations equal to one another,

F * V1 = (F + mgSin(angle)) * V2

We then solve for V2,

V2 = (F * V1) / (F + mgSin(angle))

The problem here is we don’t know what the force is i.e. what’s the force from running on a treadmill? Let’s use another equation to help.

Work = Force * Distance

-> Work and energy are the same thing.

Energy = 1/2 * mass * velocity^2

So we can say,

F * D = 1/2 * m * v^2

F = (1/2 * m * v^2) / d

We now know what F is, but we need to know what d is.

-> *This is where the theoretical aspect of things we’re not sure about!*

We are going to call d the person’s stride length. That is, the length a person generates their force over. Not the total distance ran. The idea here is while the person’s torso is moving at a constant velocity, the thing generating the movement, the legs, are not.

The stride length we are going to use is 1.50 meters. From **this **paper, 1.50 is the average between men and women, running at about 10 miles per hour.

So now that we can actually get a value for F, let’s go back to this equation,

V2 = (F * V1) / (F + mgSin(angle))

The above answer will be in meters per second. To convert to miles per hour all we do is,

V2 = 2.24 * ((F * V1) / (F + mgSin(angle))

Now a problem,

“You weight 120lbs (54kg). (Rough woman runner’s weight.) Your treadmill goes to 10.0 miles per hour. If we want to get the same workload / power output as 10 mph at 0 incline, what speed do we need to be at for other inclines?”

Here is our full equation, but don’t worry if your eyes glaze over. You can copy and paste this and just substitute actual numbers. A 5 year old can do it, I promise.

V2 = 2.2369 * (( 1/2 * m * v^2 * 1 / d ) * ( v )) / ( 1/2 * m * v^2 * 1 / d + m * 9.8 sin(angle))

m = mass

v = velocity at no incline

d = stride length

angle = up to you (between 0 and 15 degrees)

The only thing you have to do is convert your mass and velocity. For instance, 10 mph to meters per second. You can literally type in “10 mph to m/s” in google, and it gives you the answer without even hitting search. I prefer **WolframAlpha** though.

*For mass it’s 120lbs / 2.2 = kilograms*

*For velocity it’s 10 miles per hour / 2.24 = 4.47 m/s*

Next, copy and past this into Wolfram,

plot f(x) = 2.2369 * (( 1/2 * m * v^2 * 1 / d ) * ( v )) / ( 1/2 * m * v^2 * 1 / d + m * 9.8 sin(x degrees)) from x = 0 to x = 15

Then substitute mass, velocity and distance,

m = 54kg

v = 4.47

d = 1.5 meters

Deriving more specific numbers,

- Increase the incline to 2.5 and you can now run at 9.3mph
- Incline at 7.5, speed can be 8.5 mph
- Incline at 15 (typically the max for a treadmill) and speed can be 7.2 mph

These numbers match up quite well with my actual client experience.

–

**W**hat I’ve actually been looking at though is, what happens if your treadmill only goes to 10mph, and you want to get a power output above that? Where 10mph is the max V2 can be.

“You are 54kg, and you want to get the same power output of 10.5mph at 0 incline, but at 10.0mph and at what incline?”

This is what we had earlier,

V2 = (F * V1) / (F + mgSin(angle))

But now we need to solve for sin(angle),

sin(angle) = ((( F * V1) / V2) – F ) / (m * 9.8)

Substituting for F here makes things really messy. Let’s just solve for it, then plug it in,

F = (1/2 * m * v1 ^2) / d

v1 = 4.7 m/s => 10.5mph -> the speed we want to be replicating. The speed above the treadmill max.

m = 54kg

d = 1.5 m

F = 398 newtons

Back to our equation,

sin(angle) = ((( F * V1) / V2) – F ) / (m * 9.8)

angle = sin^-1(((( F * V1) / V2) – F ) / (m * 9.8))

angle = sin^-1((((398 * 4.7 ) / 4.47) – 398) / (54 * 9.8))

We could run at 2.2 degrees, at a speed of 10.0 mph, to equal our 10.5 mph at 0 incline.

Or we could do this graphically again.

plot f(x) = 2.2369 * (( 1/2 * m * v^2 * 1 / d ) * ( v )) / ( 1/2 * m * v^2 * 1 / d + m * 9.8 sin(x degrees)) from x = 0 to x = 15

We’ll say the V we’re looking to start with, our V at no incline, is 11 miles per hour / 2.24 = 4.9 m/s.

We can see at a speed of 10.0 mph, the incline would have to be 4.5 degrees, to equal the output at 11 miles per hour and no incline.

–

### Caveats to the above model

–

#### This is one way

We did this problem a few different ways. I show a couple other ways later on. Theoretically, the physics people seem to lean towards the model above, which is why I went with that first. Practically, the other models may be simpler for some to implement, the equations are shorter, and the results don’t vary much.

#### –

#### You must be running

That said, all the models break down at slower speeds. Remember how we talked about stride length earlier, or at what interval a person is generating their force. I think why these models break down at slower speeds is you’d have to adjust the stride length based on how fast the person is going.

Meaning if you start out with a speed where a person is running, but then try to use this model and extrapolate to a speed where the person is walking, it can be quite off.

For example, I have one guy who is just about jogging at nearly 3.6 miles per hour, on flat ground. But he can walk at an incline of 10 at 3.3 miles per hour.

If I use the model and say his max speed is 3.6, then see what the model gives at an incline of 10, I get this:

At an incline of 10, it says this person should only be at 1.7 miles per hour. I know that’s not right.

But then again, that’s not what this model is intended for. It’s for people who are damn near sprinting on a treadmill, and want numbers which also mean they are running on a treadmill.

Most people start jogging around 4.0 miles per hour on a treadmill. Most people start more of a run around 6.5 miles per hour. You don’t want to be using a 6.5 mph starting point, which may be trying to predict incline angles at a speed of 3.5 mph. You don’t want to be using 4.0mph as a starting point, and be trying to predict changes at 3.0mph.

–

#### Not all models work for all people

Physics and mathematical modeling can be amazing things. The ability to predict is incredible. But the surest way to screw up any ability to predict is to throw some humans into the model. The reason the stock market can’t be predicted is because you can’t predict human behavior.

You may be able to predict a good amount of a particular behavior, but that’s probably the best you can hope for.

Which is what I think this model can give. For a specific subset of people who are able to run fast on a treadmill, this will probably be right in the neighborhood. But that doesn’t mean it will work for everyone.

What if someone has never run at an incline? Well, their predicted incline might be too high initially.

What if a person has calf or achilles problems? Well, you might want to always lower how fast you run on an incline -added stress to calf / achilles complex- just to be safe.

Or what if the person hasn’t done more than a walk in a couple decades? The older guy I referenced who is at 3.6 on flat ground, but only drops to 3.3 at an incline of 10, hasn’t jogged in a very, very long time. I’m not sure he can do it at this point. He’s 78 years old, had a spinal fusion not too long ago, and has incredibly atrophied calf muscles from some nerve damage. Walking he is quite good at, which may explain why there isn’t much difference in his walking ability at inclines. Where perhaps if he didn’t have his history, his flat ground ability would be much higher, and then may line up with his incline work. I don’t know. Which is the point. Humans have flaws, which will reflect themselves in probably any model.

#### –

#### What happened to the perpendicular force?

You may have noticed we didn’t use this factor:

Some numbers, using a mass of 54kg and g is 9.8 m/s^2:

- On flat ground a person’s weight is,
- mass * gravity = 529N

- On an incline a person’s perpendicular-to-the-incline weight is,
- m*g*cos(angle)
- Using various angles,
- mgCos(5) = 527N
- mgCos(10) = 521
- mgCos(15) = 511

This means when at an incline, a person actually has a little *less* force to deal with in the perpendicular direction, in conjunction with the extra parallel force (mgSin(angle)) we went over earlier.

- mgSin(5) = 46
- mgSin(10) = 92
- mgSin(15) = 137

At each respective angle, we can see the parallel force matters a lot more than the perpendicular force.

I had a great high school physics teacher. He had his moments where he would push the class with a harder problem. Some days the class wasn’t feeling it, or wasn’t getting it. It was customary for me or another guy to raise our hand -when nobody else is raising their hand on a very hard problem- and to go, “The answer is negligible.”

It was basically an out. It’s something you’ll see in physics after a little while, where sometimes you just don’t worry about certain variables because the affect seems so small. I think it’s ok in this case to ignore the perpendicular force. For instance, from 0 degrees to a 5 degree angle, we’re not worried about a difference of 529N to 527N. Even at a max angle of 15 degrees -an angle I don’t think we’re going to need in this context- the perpendicular force only changes by ~3%. (529 changes to 511.)

Now, if you’re training a high level athlete, then this 3% may matter. However, the importance of that 3% will get lessened as this is only one component of the overall force. But again, unless you have a really slow treadmill, or a very fast runner, necessitating such a large incline, I don’t think we’ll need to worry about this. We’re mainly talking differences of 1.5% (if that) or less, and few have someone where they need to worry about this.

–

### Other models

*The following is just to show some other ways to do this. It might not be as thorough as the above. Everything that I wanted to address has been hit with the above, so the following is more optional in a sense. *

–

**W**ithout making things really complicated, and for the average person, we don’t know how much force a person is generating on a treadmill. Nor we do know how much force a treadmill is generating. As my brother said, “this is weird because the ground is moving,” this gets odd because technically the person and the treadmill are moving at a constant velocity, which means there is no acceleration. (And both the person and the treadmill are exerting a force in the same direction! Making this *really *weird.)

Force = Mass * Acceleration

Force = Mass * 0

Force = 0

But intuitively we know this isn’t correct. The person and the treadmill are each generating force. The *net* force between them is 0 though. Whatever the treadmill generates, the person (hopefully!) generates.

But the person is moving at a constant rate. What is their acceleration? What is their force? It can’t be zero, otherwise they’d fall off the treadmill.

-> *Again, this is theoretically where things may get a little dicey.*

What I think you can say here is while the person’s overall body is moving at a constant rate, the aspect of the person generating force -the legs- is not moving at a constant rate. The legs are going through cycles, accelerating and decelerating every cycle.

The torso is moving at a constant velocity, but the legs are not. What I’m going to say here is the legs are going through their cycle every second.

Velocity = Acceleration * Time

Acceleration = Velocity / Time

Because we’re going to say our time interval is one second,

Acceleration = Velocity / 1 second

Acceleration = Velocity

Going back to our other equation,

V2 = (F * V1) / (F + mgSin(angle))

F = mass * acceleration

For us -> F = mV1

V2 = (mV1 * V1) / (mV1 + mgSin(angle))

Our mass cancels out,

V2 = (V1^2) / (V1 + gSin(angle))

Now for the eyeball test.

“Let’s say we’re running at 10.0 miles per hour on a treadmill at an incline of 0 degrees. We bump the incline to 5 degrees. What velocity should we run at to have the same power output as we had with no incline?”

10 miles per hour = 4.47 m/s (unit conversion)

g = 9.8 m/s^2 = the acceleration from gravity pulling us down

(and we’ll also say acting every second)

angle = 5 degrees

V2 = (V1^2) / (V1 + gSin(angle))

V2 = (4.47^2) / (4.47 + 9.8*Sin(angle))

V2 = 3.75 m/s

3.75 * 2.24 =8.4 miles per hour

-> 2.24 is a unit conversion factor, meaning our equation could really be,

**V2 = [(V1^2) / (V1 + gSin(angle))] * 2.24**

(remember, you can merely copy and paste this into WolframAlpha, and substitute actual numbers.)

If we’re running at 10 miles per hour with no incline, then bump the incline to 5 degrees, we would want to decrease the speed to 8.4 miles per hour.

Let’s look at another person. Let’s say this person wants to generate the power of 10 miles per hour at 0 degree incline, but at 10 degree incline.

V1 = 10.0 miles per hour / 2.24 = 4.47 m/s

New incline = 10

V2 = [(4.47^2) / (4.47 + 9.8*sin(10))] * 2.24

= 7.25 miles per hour

This is right about what I’ve seen with this client. When 10.0 miles per hour was about her max speed at no incline, 7ish miles per hour was her max speed at this incline.

–

#### Another version

We’re on our treadmill with an incline of 0. Let’s say we’re moving at 10.0 mph, and we are 120lbs.

Our variables:

-m = mass = 54kg

(120lbs)

-g = acceleration due to gravity = 9.8 m/s^2

(how much gravity is pulling us down)

-Fg = mass * gravitational acceleration = 529 Newtons

(force of gravity pulling us down)

-V1 = velocity => 4.47 m/s

(converted from 10.0 mph)

Next, we’re at an incline:

-m = 54kg

-g = 9.8 m/s^2

-Gravity hasn’t changed and neither has the mass of the person

-Incline angle = 5 degrees.

(randomly chosen angle)

When at an incline we’re looking to see how much more are we being moved backward. The treadmill is always moving us backwards so long as it’s moving. When at an incline though, we are being moved backwards even more, from the incline. How much?

Giving us the variables of,

-F perpendicular = cos(5) * Fg

(force of person into the treadmill)

-F parallel = sin(5) * Fg

(force of incline moving person backwards)

The parallel force is what we’re concerned with.

-F parallel = sin(5) * Fg = 46 N

What this means is at an incline of 5 degrees, our body has to generate an extra 46N of force. How much extra speed does this translate to?

-Force = Mass * Acceleration

=> Acceleration = Force / Mass

A = 46N / 54kg = 0.85 m/s^s

We have to generate an extra 0.85 m/s^2 in acceleration when at an incline of 5 degrees, in order to not be moved backward by the incline.

We have to do this every second. So we can say,

-Velocity = Acceleration * Time

=> V = (0.85)(1) = 0.85 m/s

Using the unit conversion factor of 2.24, we have,

-0.85 m/s^2 * 2.24 = 1.9 miles per hour (per second)

Meaning if we’re at an incline of 5 degrees, we can go 1.9 miles per hour slower to equal the force it would take to move 10.0 miles per hour at a 0 degree incline.

### –

#### Tidying up

You don’t want to have to do all that if you don’t have to. It’s not tough math or anything, but if we move to variables only, we can make this cleaner.

We have our acceleration,

-Acceleration = force / mass

We have our force,

-F parallel = sin(angle) * Fg

Substituting our F parallel force into the acceleration equation,

=> Acceleration = F parallel / mass

Substituting what F parallel actually is,

=> Acceleration = (sin(angle) * mass * gravity) / mass

We can get rid of mass then as it cancels itself out,

=> Acceleration = sin(angle) * gravity

We know our body has to overcome that acceleration every second, giving us our velocity, so we can say,

=> Acceleration = sin(angle) * gravity * 1 second

=> Velocity = sin(angle) * gravity

We always have the same conversion factor to get to miles per hour (for our American treadmills),

=> V = sin(angle) * gravity * 2.24

(V here is the *extra* velocity needed at a given angle)

**V = sin(angle) * 22**

To reiterate,

V = sin(5) * 22 = 1.91 miles per hour

Smart phones have a sin function on them (iPhone = go to calculator and turn your phone sideways), so most can do this calculation wherever.

–

“If we’re running at an incline of 10 at 9.0 miles per hour, how fast would be comparable at an incline of 0?”

V = sin(10) * 22 = 3.8 miles per hour

=> 3.8 miles per hour + 9.0 miles per hour = 12.8 miles per hour

#### –

#### Going for the angle instead of the speed

“I want to run at 11 miles per hour, but our treadmill only goes to 10 miles per hour. How much of an incline do I need to be at for things to be comparable?”

We know the velocity of the treadmill will be 10.0 miles per hour, as that’s as fast as it goes.

11.0 – 10.0 = velocity difference = 1 mile per hour we need to make up.

V difference = sin(angle?) * 22

1 = sin(angle?) * 22

sin(angle?) = 1 / 22

angle = InverseSin (1 / 22)

angle = 2.6 degrees

Note that inverse sin is when sin^-1. This button,

This might not work great on your phone, but there are plenty of calculators online. If you’re using a smart phone, then you probably have internet access anyways. I like **this** calculator. (asin is inverse sin.)

This is probably the more useful way and equation,

**V difference = V what you want to be at – V treadmill**

(desired speed – actual speed)

**angle = InverseSin (V difference / 22)**

–

Let’s do one more.

“I want to haul ass, but my treadmill only goes to 10 miles per hour. What incline should I be at if I want things to be comparable to 15 miles per hour?”

V difference = 15 – 10.

angle = InverseSin (5 / 22) = 13.1 degrees

*Miscellaneous, Sports*

Sewergal2

October 28, 2015

Hey Brian, all your posts are very informative but this one was of particular interest as I foresee treadmill inclines in my future. 😀

reddyb

October 29, 2015

You may be correct 🙂

zitompul

March 2, 2018

I run 8.9 km/h on flat surface. On treadmill, for 6% inclination (3.43 degree) and the same effort (running load), I am comfortable with 7.5 km/h only. Your formula is somehow accurate in my case. Great.

b-reddy

March 2, 2018

Nice! Thank you for letting me know.