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.
Why do you want to do that?Rev. Cheeseman said:How to convert 51065 foot pound per second to Newtons, not Newtons meter per second?
Yes.Rev. Cheeseman said:1 newton is equal to 1 kg*1 meter (3 feet 3 inches) per sec per sec
No. 1 kg per sec per sec is not equal to 2 kg per sec.Rev. Cheeseman said:thus is 1 newton equal to 2 kg*1 meter per sec
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.
Still no. 1 m per sec per sec is not equal to 2 m per sec either.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.
jack action said:Why do you want to do that?Rev. Cheeseman said:How to convert 51065 foot pound per second to Newtons, not Newtons meter per second?
jack action said:Still no. 1 m per sec per sec is not equal to 2 m per sec either.
You still haven't answered this question:
Bystander said:
The highlighted text in that link only mentions forces measured in Newtons: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.
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.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?
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.
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:Moving a 51065 lbs object a foot for 1 second, is that also the same as [...]
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).Rev. Cheeseman said:So, 51065 foot pound per second can be 50 newton or 5000 newton depending on the circumstances?
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.
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.
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.
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.
andHowever, 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.
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.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.
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.