Do I actually need a blade balance?
- Thread starter BackyardBrian
- Start date
Gord Baker
Well-Known Member
So the Cone is at the Other edge. 3/16" difference from centre.. Still centre of weight distrubution. It's not the Space Shuttle per "Terryl". !
The Oregon balancer has worked well for me since I bought it about 15yrs. ago to use exclusively on my exmark Lazer Z 60". The price back then was around $90. Well made.
I live in Miami where everyone has St. Augustine grass and a sharp mower blade will make a visible difference on the blades of grass. A dull mower blade leaves a jagged cut whereas a sharp blade gives a nice clean cut across the top of the grass blade.
The problem is that the mower blades don't stay sharp beyond a month or so. So, I don't stress over it. I sharpen or replace the blades maybe twice a year. I use a flap disk and a basic cone balancer. I'll probably replace them now as when I engage the blades at idle, the whole deck starts shaking. However, at full throttle it is fine. Maybe I bent a blade and threw it out of balance.
If I could find an Oregon balance for cheap, I might buy it.
The problem is that the mower blades don't stay sharp beyond a month or so. So, I don't stress over it. I sharpen or replace the blades maybe twice a year. I use a flap disk and a basic cone balancer. I'll probably replace them now as when I engage the blades at idle, the whole deck starts shaking. However, at full throttle it is fine. Maybe I bent a blade and threw it out of balance.
If I could find an Oregon balance for cheap, I might buy it.
Tell ya what.....drive a big spike into that beam and hang a tire and wheel assembly from the lug nut/stud hole....betcha it "balances" every time.
Today's lesson in "rocket science"....
Does the force at the end of a mower blade inctease linearly with rpm
No. The (centrifugal) force at the tip of a mower blade scales with the square of RPM, not linearly.For a rotating blade the dominant structural force at the outer end is the centrifugal (outward) force that the material must resist to keep the tip moving in a circle:F=m ω2 rF = m\,\omega^{2}\,rF = m\,\omega^{2}\,r
where (m) is the effective mass of the outer portion of the blade,
(r) is the radius to the tip,
ω\omega\omega
is the angular speed in rad/s.
RPM ((N)) is simply proportional to ω\omega\omega
(ω=2πN/60\omega = 2\pi N/60\omega = 2\pi N/60
), soF∝N2.F \propto N^{2}.F \propto N^{2}.
Doubling the RPM therefore multiplies the tip force by four.(The linear tip speed v=ωrv = \omega rv = \omega r
does rise linearly with RPM, and the kinetic energy of a particle at the tip rises with v2v^{2}v^{2}
, but the force required to constrain the blade itself follows the ω2\omega^{2}\omega^{2}
relation above.)Other loads (cutting resistance, aerodynamic drag, etc.) exist, yet they are normally secondary to the centrifugal stress at high rotational speeds and still do not produce a strictly linear dependence
on RPM.
Knowing this, what would be the imbalance at the end of a 20" mower blade given a 1/8th oz higher weight at one end of the blade?
Thought for 24s
No single fixed number exists without RPM, but the unbalanced (centrifugal) force can be calculated precisely.A 20-inch blade has a 10-inch radius. An extra 1/8 oz (0.125 oz) at one tip creates a static imbalance of 1.25 oz-in.The rotating unbalanced force is:F≈1.77×(oz-in)×(RPM1000)2(in pounds)F \approx 1.77 \times (\text{oz-in}) \times \left(\frac{\text{RPM}}{1000}\right)^{2} \quad \text{(in pounds)}F \approx 1.77 \times (\text{oz-in}) \times \left(\frac{\text{RPM}}{1000}\right)^{2} \quad \text{(in pounds)}
or equivalentlyF=m ω2 rF = m\,\omega^{2}\,rF = m\,\omega^{2}\,r
(with consistent units). Because force scales with the square of speed, the effect grows quickly.Typical values for a lawn-mower bladeMost walk-behind and riding mowers run the blade in the 3000–3600 RPM range (tip speeds kept under the ANSI limit of ~19 000 ft/min).RPM
Unbalanced force
3000
≈ 20 lb
3200
≈ 23 lb
3600
≈ 29 lb
So a mere ⅛ oz difference at the tip of a 20-inch blade produces roughly 20–30 pounds of shaking force at normal operating speed. That is why even small imbalances cause noticeable vibration, premature bearing wear, and can be dangerous.If you know the exact RPM of your mower, plug it into the formula above for a more precise figure.
That's 3.54 grams......
Cleaned up....
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Gord Baker
Well-Known Member
Gord Baker
Well-Known Member
Do NOT engage your blades at idle. Belts are stretching.. Blades are NOT to be a knife edge but actually square, 0.050 vertical gives 2 cutting edges.
This Topic should be ended now.
This Topic should be ended now.
Tell ya what.....drive a big spike into that beam and hang a tire and wheel assembly from the lug nut/stud hole....betcha it "balances" every time.
Whoever "Terryl" is doesn't know crap about physics......and yes, lots of physics were used in the space shuttle.
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ILENGINE
Lawn Royalty
In @Turfspray case that would not be true. The issue isn't the sharpening grinding down the blade. The issue is the Florida sand, or in the case of Illinois the sandy/loam soils that act as a sandblaster and will cut the air lifts off the blades in a mowing season or less.If you sharpened less your blades would last much longer.Your training/theories are myth ie not based in science.