# One Arm Barbell ROws

So, in the grand scheme of things I guess it doesn’t really matter as long as form is good and muscles are being stimulated. That said, the “geek” in me can’t help but wonder what my actual weight is.

I have outgrown my largest dumbell, 90#, and have begun doing one arm rows as in this video:

When rowing like this, obviously you’re not pulling 45# of bar. What’s a good assumption for how much weight to include for the bar? 10lb, 15lb, 20lb? I admit, I have no clue.

I’m currently pulling 2X45 and 1X35 plate on this particular exercise but I’m stuck wondering what me “true” wight is.

What do you guys think?

1 pound

[quote]BONEZ217 wrote:
1 pound[/quote]

lol

OP, Why worry over the numbers at this point? You made the switch, now turn on the motivation to progress with it.

[quote]Iron Dwarf wrote:

[quote]BONEZ217 wrote:
1 pound[/quote]

lol

OP, Why worry over the numbers at this point? You made the switch, now turn on the motivation to progress with it.

[/quote]

lol yeah it really doesnt matter as long as you are progressing towards your goals

I never bothered counting the bar for this very reason.
Same with t-bar rows.

Dont forget you’re also lifting your arm. And your arm might be wearing clothes. The resistance may also be affected by the length of your fingernails. You’ve got some serious math to do fella.

The Bar does not exist.
Neither does the spoon.

2x45’s plus 1x35 should “feel” about like 150-166 lbs (the weight “feels” lighter at the top and heavier as it nears parallel to the floor).

This assumes the length of your bar L = 87", mass of bar m = 45 lb, the distance from the elevated end of the bar to where you grip it is D =20", the sleeve area where plates can be loaded is 15.5", and your plates are 1.5" thick.

Using the info above and letting:
A = angle between bar and floor
M = mass of the loaded plates
d = center of mass of plates measured from elevated end of bar
W = how heavy the weight “feels”

then you should end up with something like:

W = cos(A)*[M(L-d) + mL/2] / (L-D)

Yep, I was bored and for some strange reason felt compelled to see if I was too rusty to solve this elementary statics problem.

Feel better now OP? You’re welcome.

[quote]fizisyst wrote:
2x45’s plus 1x35 should “feel” about like 150-166 lbs (the weight “feels” lighter at the top and heavier as it nears parallel to the floor).

This assumes the length of your bar L = 87", mass of bar m = 45 lb, the distance from the elevated end of the bar to where you grip it is D =20", the sleeve area where plates can be loaded is 15.5", and your plates are 1.5" thick.

Using the info above and letting:
A = angle between bar and floor
M = mass of the loaded plates
d = center of mass of plates measured from elevated end of bar
W = how heavy the weight “feels”

then you should end up with something like:

W = cos(A)*[M(L-d) + mL/2] / (L-D)

Yep, I was bored and for some strange reason felt compelled to see if I was too rusty to solve this elementary statics problem.

Feel better now OP? You’re welcome.[/quote]

Yes!
Good Post.

This brought my nightmares of physics class back. Damn you

[quote]fizisyst wrote:
2x45’s plus 1x35 should “feel” about like 150-166 lbs (the weight “feels” lighter at the top and heavier as it nears parallel to the floor).

This assumes the length of your bar L = 87", mass of bar m = 45 lb, the distance from the elevated end of the bar to where you grip it is D =20", the sleeve area where plates can be loaded is 15.5", and your plates are 1.5" thick.

Using the info above and letting:
A = angle between bar and floor
M = mass of the loaded plates
d = center of mass of plates measured from elevated end of bar
W = how heavy the weight “feels”

then you should end up with something like:

W = cos(A)*[M(L-d) + mL/2] / (L-D)

Yep, I was bored and for some strange reason felt compelled to see if I was too rusty to solve this elementary statics problem.

Feel better now OP? You’re welcome.[/quote]

Id like to see a proof of this equation.

I still think 1lb is the right answer.

[quote]fizisyst wrote:
2x45’s plus 1x35 should “feel” about like 150-166 lbs (the weight “feels” lighter at the top and heavier as it nears parallel to the floor).

This assumes the length of your bar L = 87", mass of bar m = 45 lb, the distance from the elevated end of the bar to where you grip it is D =20", the sleeve area where plates can be loaded is 15.5", and your plates are 1.5" thick.

Using the info above and letting:
A = angle between bar and floor
M = mass of the loaded plates
d = center of mass of plates measured from elevated end of bar
W = how heavy the weight “feels”

then you should end up with something like:

W = cos(A)*[M(L-d) + mL/2] / (L-D)

Yep, I was bored and for some strange reason felt compelled to see if I was too rusty to solve this elementary statics problem.

Feel better now OP? You’re welcome.[/quote]

Wouldn’t it be sin (A)? Cos would give you the component of force that runs parallel to the ground, as least since you specified A being the angle from ground to the bar at any time.

BONEZ, sorry but no proof. I’d have post a drawing of a pretty picture and show how to setup and solve 3 equations with 3 unknowns. Too much work!

Fisch, it wouldn’t be sin(A). If you take the simpler case neglecting the angle A and assuming the bar stays parallel to the floor, then you just get: W = [M(L-d) + mL/2] / (L-D). When the bar is parallel to the floor, A=0, cos(0) = 1, and W is at a max.

If you lifted the bar all the way until it’s sticking straight up in the air then A=90 and cos(90) = W = 0. At this point the bar is balanced on one end (albeit unstable!) and you’re not having to fight gravity at all.

That solution assumes that your muscles only resist the weight in the vertical vector however, correct?

Sharp, the solution assumes your arm is always pulling perpendicular to the bar. So at the bottom you’re pulling almost vertically, but at the top you’re pulling slightly back at an angle.

The forearm is like a rope attached to the bar, and we’re trying to find the tension in the rope. The tension vector continuously changes direction as the bar moves, always remaining perpendicular to the bar. Any component of force applied in any other direction is wasted effort, since the bar moves in an arc.

Errm…me pick fings up and put dem down!

Sooooo, thats the equation, whats the actual answer in a number?

just sayin

[quote]late2thegame wrote:
So, in the grand scheme of things I guess it doesn’t really matter as long as form is good and muscles are being stimulated. That said, the “geek” in me can’t help but wonder what my actual weight is.

I have outgrown my largest dumbell, 90#, and have begun doing one arm rows as in this video:

When rowing like this, obviously you’re not pulling 45# of bar. What’s a good assumption for how much weight to include for the bar? 10lb, 15lb, 20lb? I admit, I have no clue.

I’m currently pulling 2X45 and 1X35 plate on this particular exercise but I’m stuck wondering what me “true” wight is.

What do you guys think?[/quote]
You would need vector math to figure this out.

[quote]Tech-Junkie wrote:

[quote]late2thegame wrote:
So, in the grand scheme of things I guess it doesn’t really matter as long as form is good and muscles are being stimulated. That said, the “geek” in me can’t help but wonder what my actual weight is.

I have outgrown my largest dumbell, 90#, and have begun doing one arm rows as in this video:

When rowing like this, obviously you’re not pulling 45# of bar. What’s a good assumption for how much weight to include for the bar? 10lb, 15lb, 20lb? I admit, I have no clue.

I’m currently pulling 2X45 and 1X35 plate on this particular exercise but I’m stuck wondering what me “true” wight is.

What do you guys think?[/quote]
You would need vector math to figure this out.
[/quote]

No you dont. The answer is clearly 1lb.

[quote]BONEZ217 wrote:

[quote]Tech-Junkie wrote:

[quote]late2thegame wrote:
So, in the grand scheme of things I guess it doesn’t really matter as long as form is good and muscles are being stimulated. That said, the “geek” in me can’t help but wonder what my actual weight is.

I have outgrown my largest dumbell, 90#, and have begun doing one arm rows as in this video:

When rowing like this, obviously you’re not pulling 45# of bar. What’s a good assumption for how much weight to include for the bar? 10lb, 15lb, 20lb? I admit, I have no clue.

I’m currently pulling 2X45 and 1X35 plate on this particular exercise but I’m stuck wondering what me “true” wight is.

What do you guys think?[/quote]
You would need vector math to figure this out.
[/quote]

No you dont. The answer is clearly 1lb.[/quote]

ergo it doesn’t matter.

you have 125 lb on the bar and that’s a number you can add to and progress with.

everything else is just a detail that is unimportant in the big scheme of things.

[quote]fr0IVIan wrote:

[quote]BONEZ217 wrote:

[quote]Tech-Junkie wrote:

[quote]late2thegame wrote:
So, in the grand scheme of things I guess it doesn’t really matter as long as form is good and muscles are being stimulated. That said, the “geek” in me can’t help but wonder what my actual weight is.

I have outgrown my largest dumbell, 90#, and have begun doing one arm rows as in this video:

When rowing like this, obviously you’re not pulling 45# of bar. What’s a good assumption for how much weight to include for the bar? 10lb, 15lb, 20lb? I admit, I have no clue.

I’m currently pulling 2X45 and 1X35 plate on this particular exercise but I’m stuck wondering what me “true” wight is.

What do you guys think?[/quote]
You would need vector math to figure this out.
[/quote]

No you dont. The answer is clearly 1lb.[/quote]

ergo it doesn’t matter.

you have 125 lb on the bar and that’s a number you can add to and progress with.

everything else is just a detail that is unimportant in the big scheme of things.
[/quote]

lol im just messing with the math nerds. lots of things are unimportant in the big scheme, some people still enjoy the quest for knowledge.