# How to convert ft-lbs/sec to Newtons?

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• Rev. Cheeseman
Rev. Cheeseman
Hi,

How to convert 51065 foot pound per second to Newtons, not Newtons meter per second?

Are the dimensions of both units consistent? ft-lb appears to be a unit of energy, so the equivalent SI unit is the Joule, or equivalently the Newton meter. ft-lb per second is then a unit of power, for which the equivalent SI unit is the Watt, or equivalently the Newton meter per second.

pasmith said:
Are the dimensions of both units consistent? ft-lb appears to be a unit of energy, so the equivalent SI unit is the Joule, or equivalently the Newton meter. ft-lb per second is then a unit of power, for which the equivalent SI unit is the Watt, or equivalently the Newton meter per second.

Ok, correct me if I'm wrong. 1 newton is equal to 1 kg*1 meter (3 feet 3 inches) per sec per sec, thus is 1 newton equal to 2 kg*1 meter per sec or 46 lbs at 3 feet 3 inches in 1 second?

Or is 1 Newton equals to 1 kg 2 meters per 1 second?

Using an online converter, 51064 foot pounds per second (ft lbf/s of power) is equals to 69 233.49 Newtons meter/second (N m/s / power). Converting it to kg meter/sec, it will be 7059 kg*1 meter per sec. 7059 kg*1 meter per sec convert to pound meter per sec will be 15562 lbs at 1 meter (3 feet 3 inches) per sec. Is it appropriate?

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Rev. Cheeseman said:
How to convert 51065 foot pound per second to Newtons, not Newtons meter per second?
Why do you want to do that?

Rev. Cheeseman said:
1 newton is equal to 1 kg*1 meter (3 feet 3 inches) per sec per sec
Yes.

Rev. Cheeseman said:
thus is 1 newton equal to 2 kg*1 meter per sec
No. 1 kg per sec per sec is not equal to 2 kg per sec.

Get familiar with dimensional analysis and the factor–label method to make the proper conversion.

jack action said:
Why do you want to do that?

Yes.

No. 1 kg per sec per sec is not equal to 2 kg per sec.

Get familiar with dimensional analysis and the factor–label method to make the proper conversion.

Sorry the 2kg 1 m 1 sec is a typo, it should be typed as 1 kg 2 m 1 sec. Is that correct? Sorry.

I want to compare the 51065 ft lbs per sec figure to another after it is converted to Newtons. I tried to find online converters where we can convert foot pound per sec figures to Newtons but I only managed to find foot pound per sec converted to Newtons meter per sec.

First of all, my background in mathematics is very limited.

Rev. Cheeseman said:
Sorry the 2kg 1 m 1 sec is a typo, it should be typed as 1 kg 2 m 1 sec. Is that correct? Sorry.
Still no. 1 m per sec per sec is not equal to 2 m per sec either.

You still haven't answered this question:
jack action said:
Rev. Cheeseman said:
How to convert 51065 foot pound per second to Newtons, not Newtons meter per second?
Why do you want to do that?

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Rev. Cheeseman said:
I want to compare 51065 foot pound per second to Newton so I can compare its impact force to these punches and kicks here in Newtons https://www.mdpi.com/2075-4663/12/3/74#:~:text=In terms of impact force,were reported at 122.6 N.
The highlighted text in that link only mentions forces measured in Newtons:
"In terms of impact force, roundhouse kicks’ impact force ranged from 172.0 to 6400 N; side kicks ranged from 461.8 to 9015 N; back kicks ranged from 562.4 to 8569 N; front kicks ranged from 466.6 to 7790 N; and axe kicks were reported at 122.6 N."

Could you explain the origin of that 51065 foot pound per second?

Rev. Cheeseman said:
51065 foot pound per second is basically the capability to move 51065 lbs object a foot in one second, isn't it?
It is actually a 51065 lb force that is maintained for 1 second over a distance of 1 foot. So elevating a 51065 lb object, against gravity, at constant velocity (no acceleration), by 1 foot, within 1 second would be an example. The amount of energy is a force multiplied by the distance traveled and the power required is the amount of energy divided by the time taken. Your number represents a power and you are looking for a force. There is no direct conversion because there is missing information.

To compare with your punch force, you would need to know how far that punch force was maintained and how long it lasted. You would then compare power values.

Or, you would need to know how long that power output would be maintained and how far it traveled. You would then compare force values.

Note also that the distance traveled ##d## per unit of time ##\Delta t## is the average speed ##v##. So the average power output ##P## divided by the average speed ##v## of the event would give you the average force ##F## during that event.
$$P = \frac{F \times d}{\Delta t} = F \times \frac{d}{\Delta t} = F \times v$$
To "convert" one to the other, you need to know the speed at which that force is moving.

Lnewqban said:
The highlighted text in that link only mentions forces measured in Newtons:
"In terms of impact force, roundhouse kicks’ impact force ranged from 172.0 to 6400 N; side kicks ranged from 461.8 to 9015 N; back kicks ranged from 562.4 to 8569 N; front kicks ranged from 466.6 to 7790 N; and axe kicks were reported at 122.6 N."

Could you explain the origin of that 51065 foot pound per second?

jack action said:
It is actually a 51065 lb force that is maintained for 1 second over a distance of 1 foot. So elevating a 51065 lb object, against gravity, at constant velocity (no acceleration), by 1 foot, within 1 second would be an example. The amount of energy is a force multiplied by the distance traveled and the power required is the amount of energy divided by the time taken. Your number represents a power and you are looking for a force. There is no direct conversion because there is missing information.

To compare with your punch force, you would need to know how far that punch force was maintained and how long it lasted. You would then compare power values.

Or, you would need to know how long that power output would be maintained and how far it traveled. You would then compare force values.

Note also that the distance traveled ##d## per unit of time ##\Delta t## is the average speed ##v##. So the average power output ##P## divided by the average speed ##v## of the event would give you the average force ##F## during that event.
$$P = \frac{F \times d}{\Delta t} = F \times \frac{d}{\Delta t} = F \times v$$
To "convert" one to the other, you need to know the speed at which that force is moving.

Moving a 51065 lbs object a foot for 1 second, is that also the same as "a 51065 lb force that is maintained for 1 second over a distance of 1 foot. So elevating a 51065 lb object, against gravity, at constant velocity (no acceleration), by 1 foot, within 1 second would be an example"? Sorry for asking this type of question, because English is not my first language.

So, 51065 foot pound per second can be 50 newton or 5000 newton depending on the circumstances?

Rev. Cheeseman said:
Moving a 51065 lbs object a foot for 1 second, is that also the same as [...]
Maybe not. For example, if you push a 51065 lb object resting on ice, the moving force is the friction force. Since ice is very slippery, the friction force is around 1500 lb (friction coefficient of 0.03). If it was a wood object on a wood surface, the force would be about 10 times greater.

Rev. Cheeseman said:
So, 51065 foot pound per second can be 50 newton or 5000 newton depending on the circumstances?
Absolutely. With 51065 lb.ft/s of power, you can produce a force of 51065 lb moving at 1 ft/s or a 1 lb force moving at 51065 ft/s. How long or how far you can maintain that force depends on how much energy you have (in foot-pound).

Imagine a car with an engine that can transform heat energy from burning fuel to mechanical energy at a rate of 51065 lb.ft/s of power. With the gearbox, you can set the desired speed of the car. You can find the force that propels the car with those two values. How far you will go depends only on how much fuel you have in the vehicle, which is your energy source.

jack action said:
Maybe not. For example, if you push a 51065 lb object resting on ice, the moving force is the friction force. Since ice is very slippery, the friction force is around 1500 lb (friction coefficient of 0.03). If it was a wood object on a wood surface, the force would be about 10 times greater.

In this context involving strikes like punches and kicks, isn't it pushing a 51065 object a foot at one second more appropriate than lifting a 51065 lb object, against gravity, at constant velocity (no acceleration), by 1 foot, within 1 second?

I'm sorry if that sounds ignorant but we understand punches and kicks are closer to pushes than pulls.
jack action said:
Absolutely. With 51065 lb.ft/s of power, you can produce a force of 51065 lb moving at 1 ft/s or a 1 lb force moving at 51065 ft/s. How long or how far you can maintain that force depends on how much energy you have (in foot-pound).

Imagine a car with an engine that can transform heat energy from burning fuel to mechanical energy at a rate of 51065 lb.ft/s of power. With the gearbox, you can set the desired speed of the car. You can find the force that propels the car with those two values. How far you will go depends only on how much fuel you have in the vehicle, which is your energy source.

Sorry, but how to use the same explanation in the context of punching and kicking?

Can we guesstimate the speed from the timestamp 1:15 from this video

The machine in the video doesn't use typical units, so it is hard to understand what they measure.

Assuming the power output (##51064\ lb.ft/s##) and the reaction time (##0.13224\ s##) are right I found this human punch calculator that could help you. Not only it is a calculator but the equations used are all described.

The force ##F## is related to the mass ##m## of the boxer and the accelaration of the punch ##a##:
$$F= ma$$
Where the acceleration is related to the maximum speed ##v## of the punch and the time ##t## it takes to decelerate to a stop:
$$F= m\frac{v}{t}$$
As we said earlier, the power is the force times the average velocity ##v_{avg}##:
$$P = Fv_{avg}$$
Since the initial velocity is the maximum velocity of the punch and the final velocity is zero, then ##v_{avg} = \frac{v}{2}##. Thus:
$$P = \left( m\frac{v}{t} \right) \left( \frac{v}{2} \right) = \frac{\frac{1}{2}mv^2}{t}$$
Which corresponds to the kinetic energy of the boxer's body released during the reaction time period. Seems logical.

Let's find out the initial speed of the punch:
$$v = \sqrt{\frac{2Pt}{\left( \frac{W}{g} \right)}} = \sqrt{\frac{2(51064\ lb.ft/s)(0.13224\ s)}{\left( \frac{250\ lb}{32.174\ ft/s^2} \right)}} = 41.7\ ft/s$$
Here I use ##W = mg## to be able to use the weight ##W## of the boxer (I guessed a ##250\ lb## value). ##g## is the acceleration due to gravity. According to this source, ##41.7\ ft/s## (##28\ mph##) is a reasonable assumption:

Welterweight boxing world champion Ricky Hatton tested his punches at the University of Manchester. His average punch traveled at around 25 mph, while his fastest punch clocked 32 mph.

[...]

The record-breaking punch that’s currently the fastest punch ever recorded traveled at 45 mph. That’s incredible, considering that some of the fastest boxing hitters barely reach 35 mph.

Therefore:
$$F = \left(\frac{W}{g}\right)\frac{v}{t} = \left(\frac{250\ lb}{32.174\ ft/s^2}\right)\frac{41.7\ ft/s}{0.13224\ s} = 2450\ lb \equiv 10899\ N$$
Which is the result you would get with the human punch calculator presented above.

This is how I would "convert" 51064 foot-pounds per second to Newtons in this particular scenario.

Lnewqban
jack action said:
The machine in the video doesn't use typical units, so it is hard to understand what they measure.

Assuming the power output (##51064\ lb.ft/s##) and the reaction time (##0.13224\ s##) are right I found this human punch calculator that could help you. Not only it is a calculator but the equations used are all described.

The force ##F## is related to the mass ##m## of the boxer and the accelaration of the punch ##a##:
$$F= ma$$
Where the acceleration is related to the maximum speed ##v## of the punch and the time ##t## it takes to decelerate to a stop:
$$F= m\frac{v}{t}$$
As we said earlier, the power is the force times the average velocity ##v_{avg}##:
$$P = Fv_{avg}$$
Since the initial velocity is the maximum velocity of the punch and the final velocity is zero, then ##v_{avg} = \frac{v}{2}##. Thus:
$$P = \left( m\frac{v}{t} \right) \left( \frac{v}{2} \right) = \frac{\frac{1}{2}mv^2}{t}$$
Which corresponds to the kinetic energy of the boxer's body released during the reaction time period. Seems logical.

Let's find out the initial speed of the punch:
$$v = \sqrt{\frac{2Pt}{\left( \frac{W}{g} \right)}} = \sqrt{\frac{2(51064\ lb.ft/s)(0.13224\ s)}{\left( \frac{250\ lb}{32.174\ ft/s^2} \right)}} = 41.7\ ft/s$$
Here I use ##W = mg## to be able to use the weight ##W## of the boxer (I guessed a ##250\ lb## value). ##g## is the acceleration due to gravity. According to this source, ##41.7\ ft/s## (##28\ mph##) is a reasonable assumption:

Therefore:
$$F = \left(\frac{W}{g}\right)\frac{v}{t} = \left(\frac{250\ lb}{32.174\ ft/s^2}\right)\frac{41.7\ ft/s}{0.13224\ s} = 2450\ lb \equiv 10899\ N$$
Which is the result you would get with the human punch calculator presented above.

This is how I would "convert" 51064 foot-pounds per second to Newtons in this particular scenario.

Sorry, language barrier issue. Is the 'reaction time' which is 132.24 sec or 2.2 minutes the same as the 'the speed at which that force is moving'?

It looks like the 'reaction time' on the Powerkube is the time someone take to throw a strike into the pad as soon as the bell rang as we can see here,

...so the speed of the strike of the previous fighter with 51064 foot-pounds per second can only be estimated or guessed because it looks like the fighter take 2 minutes to throw the punch into the pad. Nevermind, it's fine.

Thank you so much for showing the details and formulas. Really appreciated it.

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For sure, the "reaction time" on the machine is 132.24 milliseconds (ms) or 0.13224 seconds.

You may be right as to what it measures because I really don't know how this machine works. (I couldn't find a manual online.)

Nevertheless, I thought that around 0.1 s was a reasonable estimate for "delivery time" so I thought "reaction time" from the machine was maybe just the way the makers chose to name it. I did not know about the bell and you might be right about what it represents.

But the math I presented is still good ... if the machine could give the "delivery time".

There is another way to find the time with the "compression energy" given by the machine as "535 cal X 10" which I'm assuming is 5350 calories or about 16531 lb.ft. Again, the makers of this machine use weird nomenclature and units so it is difficult to understand what is what. If this is the energy absorbed by the machine, corresponding to the work ##T## done by the punch, then:
$$P = \frac{T}{t}$$
or
$$t= \frac{16531\ lb.ft}{51064\ lb.ft/s} = 0.3237\ s$$
Note that the calculator uses ##0.3\ s## as a default value; but it seems to be an average value, not one for a highly trained boxer.

Using that value in the previous equations instead of ##0.13224\ s##, you get a speed of ##65\ ft/s## (##45\ mph##) which corresponds to the fastest punch ever recorded according to my source previously shown. The punch force becomes ##1564\ lb## or ##6955\ N##.

According to my previous source:
However, there were some studies made that calculated the pure force behind a punch in Newtons. One particular study indicated that amateur boxers generate around 2500 N with their punches.

Of course, amateur boxers and elite fighters are more skilled and stronger than your average person, but not as strong as elite-level fighters like professional boxing world champions. Elite heavyweights can generate up to 5000 N of force with their punches. That’s almost like getting hit with a sledgehammer.
and
you get 3104 pounds of force behind an average amateur boxer’s punch. That average is lower (around 1800 lbs) in the featherweight and higher (around 4264 lbs) in the heavyweight division.
which seems to contradict itself. I didn't notice that discrepancy earlier. It seems the wrong unit has been used in one of the statements and, with the context, it seems the numbers in pounds should be more realistic. But I don't know for sure.

So I have two answers for you: ##6955\ N## or ##10899\ N##. We just need to understand what the machine measures exactly.

Rev. Cheeseman
jack action said:
For sure, the "reaction time" on the machine is 132.24 milliseconds (ms) or 0.13224 seconds.

You may be right as to what it measures because I really don't know how this machine works. (I couldn't find a manual online.)

Nevertheless, I thought that around 0.1 s was a reasonable estimate for "delivery time" so I thought "reaction time" from the machine was maybe just the way the makers chose to name it. I did not know about the bell and you might be right about what it represents.

But the math I presented is still good ... if the machine could give the "delivery time".

There is another way to find the time with the "compression energy" given by the machine as "535 cal X 10" which I'm assuming is 5350 calories or about 16531 lb.ft. Again, the makers of this machine use weird nomenclature and units so it is difficult to understand what is what. If this is the energy absorbed by the machine, corresponding to the work ##T## done by the punch, then:
$$P = \frac{T}{t}$$
or
$$t= \frac{16531\ lb.ft}{51064\ lb.ft/s} = 0.3237\ s$$
Note that the calculator uses ##0.3\ s## as a default value; but it seems to be an average value, not one for a highly trained boxer.

Using that value in the previous equations instead of ##0.13224\ s##, you get a speed of ##65\ ft/s## (##45\ mph##) which corresponds to the fastest punch ever recorded according to my source previously shown. The punch force becomes ##1564\ lb## or ##6955\ N##.

According to my previous source:

and

which seems to contradict itself. I didn't notice that discrepancy earlier. It seems the wrong unit has been used in one of the statements and, with the context, it seems the numbers in pounds should be more realistic. But I don't know for sure.

So I have two answers for you: ##6955\ N## or ##10899\ N##. We just need to understand what the machine measures exactly.

Hmm... Turns out their value of force known as Franklin as we can see in these videos is invented by the inventor of the measuring device, Kevin Franklin. It is a non-standard force value so I (not just me but everyone except the inventor and the company who made the measuring device) have no idea at all how to translate it to Newton. Not to mention they don't even want to show to the public their own formula of their own force value according to one of the owner of the measuring device on YouTube. Strange.

By the way, the name of the measuring device is PowerKube. This is a study regarding the reliability of that measuring device https://www.ncbi.nlm.nih.gov/pmc/articles/PMC9784821/. It said, "For each strike, The PowerKubeTM software provided four different peak strike values: impact power (W); impact power (ft.lbs/s); kinetic energy (cal × 10) and the power index (no units). The power index and kinetic energy outcome measures are unitless, proprietary calculations". It seems like they're using an older version of the PowerKube in the study. The latest version of PowerKube uses Joules for kinetic energy instead of cal x 10. https://www.powerkube.tech/pages/the-science

I just tried to see if I can use the formula in order to see the comparison between the result of using the formula and the British heavyweight boxer Frank Bruno's stats here https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1419171/pdf/bmjcred00479-0016.pdf

By using Google, Frank Bruno's weight is approximately 250 lbs. So, using his weight and some numbers from the stats we got...

(250 lbs/32.174 ft per sec^2) 29.3 ft per sec/0.1 sec = 2276 lbs force thus 10124 N

But the real Newton value, as we can see in the stats, is actually much lower at 4096 N or 920 lbs force.

Guesstimating is all fun and I really like doing this, but it looks like if we want to get exact result especially if we want to know the the real force value of the previous fighter in my main question, we need him to punch a measuring device that gives us the value in Newton. Not a self-made force value that used non-standard unit.

However, I'm still fascinated by the way you comes up with formulas on how to convert power to force that some others think are time consuming. Very interesting. Thank you.

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jack action

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